Answer:
[tex]Sn(CO_3)_2[/tex] and [tex]NaNO_3[/tex].
Explanation:
Hello there!
In this case, according to the described chemical reaction between sodium carbonate and tin nitrate, we can write the undergoing chemical reaction:
[tex]2Na_2CO_3+Sn(NO_3)_4\rightarrow Sn(CO_3)_2+4NaNO_3[/tex]
Thus, we notice that the products are strontium carbonate and sodium nitrate, whose formulas are [tex]Sn(CO_3)_2[/tex] and [tex]NaNO_3[/tex] respectively.
Best regards!
A strong acid, such as hydrochloric acid cannot be poured down a sink because it will react and dissolve the metal in the pipes. Yet a strong base, commonly found in drain cleaner, can be poured down a sink. A strong acid can be neutralized with a strong base, creating a salty water solution. What will happen when 10.0 g of hydrochloric acid were mixed with 10.5 grams of calcium hydroxide?
Answer:
15.2 grams of calcium chloride are produced and HCl is the limiting reactant.
Explanation:
Hello there!
In this case, according to the described scenario, it is possible to realize that the reaction between hydrochloric acid and calcium hydroxide is:
[tex]2HCl+Ca(OH)_2\rightarrow CaCl_2+2H_2O[/tex]
Whereas there is a 2:1 mole ratio of the acid to the base. In such a way, with the given masses, we can compute how much calcium chloride product is produced due to the chemical reaction via stoichiometry:
[tex]m_{CaCl_2}^{by HCl}=10.0gHCl*\frac{1molHCl}{36.46gHCl}*\frac{1molCaCl_2}{2molHCl} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.2gCaCl_2\\\\m_{CaCl_2}^{by Ca(OH)_2}=10.5gHCl*\frac{1molCa(OH)_2}{74.09gCa(OH)_2}*\frac{1molCaCl_2}{1molCa(OH)_2} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.7gCaCl_2[/tex]
Whereas we infer that the correct amount is 15.2 g since HCl is the limiting reactant as it produces the fewest grams of the desired product. Consequently, the calcium hydroxide is the excess reactant here.
Regards!
Atoms in the same PERIOD have the same...
A. Number of energy levels
B. Size
C. Temperature
D. Reactivity
Answer:
A. Number of energy levels
Part A
A solution of cough syrup contains 5.00% active ingredient by volume. If the total volume of the bottle is 68.0 mL, how many milliliters of ac
ingredient are in the bottle?
Express your answer with the appropriate units.
There are 3.4 milliliters of the active ingredient in the cough syrup bottle.
To calculate the volume of the active ingredient in the cough syrup bottle, we need to multiply the total volume of the bottle by the percentage of the active ingredient.
Given:
Total volume of the bottle = 68.0 mL
Percentage of active ingredient = 5.00%
First, we convert the percentage to a decimal by dividing it by 100:
Percentage of active ingredient = 5.00% = 5.00/100 = 0.05
Next, we calculate the volume of the active ingredient:Volume of active ingredient = Total volume of the bottle × Percentage of active ingredient
Volume of active ingredient = 68.0 mL × 0.05
Volume of active ingredient = 3.4 mL
Therefore, there are 3.4 milliliters of the active ingredient in the cough syrup bottle.
It's important to note that the calculation assumes a homogeneous distribution of the active ingredient throughout the solution.
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It took 14.50 mL of 0.455M NaOH to fully neutralize 12.0mL of HCl. What is the concentration of the HCl?
HCl + NaOH \rightarrow→ NaCl + H2O
Answer:
0.550 M HCl
Explanation:
M1V1 = M2V2
M1 = 0.455 M NaOH
V1 = 14.50 mL NaOH
M2 = ?
V2 = 12.0 mL HCl
Solve for M2 --> M2 = M1V1/V2
M2 = (0.455 M)(14.50 mL) / (12.0 mL) = 0.550 M HCl
Answer:
The appropriate answer is "0.549 M".
Explanation:
The given values are:
N₁ = 14.50 mL
V₁ = 0.455 M
N₂ = 12 mL
Let
V₂ = C = ?
As we know,
⇒ [tex]N_1\times V_1=N_2\times V_2[/tex]
On substituting the values, we get
⇒ [tex]14.50\times 0.455 = 12\times C[/tex]
⇒ [tex]6.5975=12\times C[/tex]
⇒ [tex]C=\frac{6.5975}{12}[/tex]
⇒ [tex]=0.549 \ M[/tex]
Can I get some help for these chemistry questions .
1. C6H10O5 + 6O2 → 6CO2 + 5H2O
2. MgO + H2O → Mg(OH)2
3. 151.2444 grams of carbon dioxide are required to produce 3.44 moles of carbonic acid in the blood.
1. The complete, balanced chemical reaction for the burning of cellulose (C6H10O5), or wood, is:
C6H10O5 + 6O2 → 6CO2 + 5H2O
This reaction represents the combustion of cellulose in the presence of oxygen, resulting in the formation of carbon dioxide (CO2) and water (H2O).
2. The balanced equation for the reaction between magnesium oxide (MgO) and water (H2O) to produce magnesium hydroxide (Mg(OH)2) is:
MgO + H2O → Mg(OH)2
According to the equation, 1 mole of magnesium oxide reacts with 1 mole of water to produce 1 mole of magnesium hydroxide. Therefore, if 2.55 moles of magnesium oxide react, the same number of moles of magnesium hydroxide will be produced.
3. The balanced equation for the reaction between carbon dioxide (CO2) and water (H2O) to form carbonic acid (H2CO3) is:
CO2 + H2O → H2CO3
According to the equation, 1 mole of carbon dioxide reacts with 1 mole of water to produce 1 mole of carbonic acid. Therefore, if 3.44 moles of carbonic acid are produced, the same number of moles of carbon dioxide is required.
To calculate the mass of carbon dioxide, we need to know its molar mass, which is approximately 44.01 g/mol. Therefore, the mass of carbon dioxide required to produce 3.44 moles of carbonic acid is:
Mass = Moles × Molar mass
Mass = 3.44 moles × 44.01 g/mol
Mass = 151.2444 g
Therefore, 151.2444 grams of carbon dioxide are required to produce 3.44 moles of carbonic acid in the blood.
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A solution contains 1.817 mg of CoSO4 (155.0 grams/mole) per mL. Calculate the volume (in mL) of 0.009795 M Zn2 needed to titrate the excess complexing reagent after the addition of 70.00 mL of 0.009005 M EDTA to a 20.00 mL aliquot of the Co2 solution.
Answer:
85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .
Explanation:
Lets calculate
After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml
As the number of moles of [tex]CoSO_4[/tex] =[tex]\frac{Given mass }{molar mass}[/tex]
=[tex]\frac{1.817}{155}[/tex]
=0.01172
Molarity = [tex]\frac{no. of moles}{volume of solution}[/tex]
=[tex]\frac{0.01172}{20}[/tex]
=0.000586 moles
Excess of EDTA = concentration of EDTA - concentration of CoSO4
= 0.009005 - 0.000586
= 0.008419 M
As M1V1 ( Excess of EDTA ) = M2V2 [tex](Zn^2^+)[/tex]
[tex]0.008419\times100ml=0.009795\times V2[/tex]
[tex]V2=\frac{0.008419\times100}{0.009795}[/tex]
V2 =85.952 ml
Therefore , 85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .
Make the arbitrary assignment of the reduction of Zn2+(aq) to Zn(s) as you 0 V for your brief list.
Zn^2+(aq) +2e- → Zn(s) E = 0 V
The arbitrary assignment of the reduction of Zn2+(aq) to Zn(s) as E = 0 V serves as a reference point for determining the relative reduction potentials of other redox reactions. This assignment is based on the convention that the standard reduction potential
In the case of the Zn2+(aq) to Zn(s) reduction, the reaction involves the gain of two electrons by Zn2+ ions, leading to the formation of solid zinc metal. The assigned reduction potential of 0 V indicates that, under standard conditions (1 M concentration, 25°C, and 1 atm pressure), the Zn2+ ions have a tendency to accept electrons and be reduced to Zn metal.
Any reduction potential above 0 V suggests a greater tendency for reduction, while a negative reduction potential indicates a lower tendency for reduction compared to the Zn2+(aq) to Zn(s) reaction.
This reference potential allows us to compare the reactivity of other redox systems and predict the feasibility of different reactions. The more positive the reduction potential, the greater the tendency for reduction to occur. Therefore, if we encounter a reduction potential of +0.34 V for another reaction, we can infer that it is more likely to occur spontaneously compared to the Zn2+(aq) to Zn(s) reduction. Conversely, if we encounter a reduction potential of -0.50 V, we can conclude that the reverse reaction (oxidation) is more favorable than the reduction of Zn2+(aq) to Zn(s).
Overall, the assignment of E = 0 V for the reduction of Zn2+(aq) to Zn(s) provides a benchmark for understanding the electrochemical behavior of other redox reactions and allows us to make predictions based on the relative reduction potentials.
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7. A 795.0 mL volume of hydrogen gas is collected at 23 oC and 1055 torr. What volume will it occupy at STP?
The volume of hydrogen gas at STP is 670.7 mL.
The given information is,Volume of hydrogen gas = 795.0 mLTemperature (T) = 23 oCPressure (P) = 1055 torrWe are required to find the volume of hydrogen gas at STP.STP stands for Standard Temperature and Pressure. It is used as a standard for measurement of gas volume and pressure. The standard temperature is 0 oC or 273.15 K and the standard pressure is 1 atm or 760 mmHg or 101.325 kPa.1 atm = 760 mmHg = 101.325 kPaSTP Conditions:T = 273.15 KP = 1 atmFrom the ideal gas law, we havePV = nRTWhere, P is pressureV is volumeT is temperature n is the number of moles of gasR is the gas constantFor the comparison of volumes, we must have the same value of n and R on both sides of the equation.So, we can write the above equation asP₁V₁/T₁ = nR/P₂V₂/T₂Where,P₁, V₁ and T₁ are the initial valuesP₂, V₂ and T₂ are the final valuesFor the given data, we can write1055 torr x 795.0 mL / (273.15+23) K = nR/1 atm x V₂/273.15 KThe initial pressure and volume do not match the standard conditions. We have to convert the given pressure and volume to the standard conditions.Using the ideal gas law, we can writePV = nRTWe can rewrite this equation asP₁V₁/T₁ = P₂V₂/T₂Where,P₁, V₁ and T₁ are the initial valuesP₂, V₂ and T₂ are the final valuesFor the given data, we can write1055 torr x 795.0 mL / (273.15+23) K = P₂ x V₂ / (273.15 K)Rearranging this equation, we getP₂V₂ = 1 atm x 795.0 mL / (273.15 K) x 1055 torrP₂V₂ = 768.47 mLWe can now use the final pressure and volume to calculate the volume at STP.P₁V₁/T₁ = P₂V₂/T₂V₂ = P₁V₁T₂ / T₁P₁ = 1 atmV₁ = 768.47 mLT₁ = 23 oC + 273.15 = 296.15 KV₂ = 1 atm x 768.47 mL / 1055 torr x 296.15 K / 273.15 KV₂ = 670.7 mL
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Element A has two isotopes. The first isotope is present 18.18% of the time and has a mass of
147.99. The second isotope has a mass of 127.76. Calculate the atomic mass of element A. (To two
decimals places)
The atomic mass of element A, given that the first isotope has abundance of 18.18% and a mass of 147.99, is 131.43 amu
How do i determine the atomic mass of element A?From the question given above, the following data were obtained:
Abundance of 1st isotope (1st%) = 18.18%Mass of 1st isotope = 147.99Mass of 2nd isotope = 127.76 Abundance of 2nd isotope (2nd%) = 100 - 18.18 = 81.82%Atomic mass of element A=?The atomic mass of the element A can be obtain as illustrated below:
Atomic mass = [(Mass of 1st × 1st%) / 100] + [(Mass of 2nd × 2nd%) / 100]
Inputting the given parameters, we have:
Atomic mass = [(147.99 × 18.18) / 100] + [(127.76 × 81.82) / 100]
Atomic mass = 26.90 + 104.53
Atomic mass = 131.43 amu
Thus, the atomic mass of element A obtained from the above calaculation is 131.43 amu
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By examining data from distant stars, astronomers can determine
if a star is moving away from or toward Earth. Which of the
following pieces of data would be most helpful in determining
the motion of a star?
A The star gives off blue-white light.
B
The star gives off mainly radio waves and X-rays.
C
The light spectrum given off by the star is shifted toward the
red end.
.
D
The surface temperature of the star is approximately 10,000°
Celsius.
Answer:
The answer is A ahahahha
Explanation:
Because it just is
Assertion: The conversion of a gas directly into solid is called condensation. Reason : Naphthalene leaves no residue when kept in open for some time. *
The reason is not an explanation of the assertion as we can see, the concepts are not related.
What is condensation?The process by which a substance transforms from a gaseous state to a liquid state is known as condensation. When a gas's temperature is dropped, the gas molecules experience energy loss and coalesce to create a liquid.
The kinetic energy of gas molecules is transformed into potential energy during condensation when they slow down and move closer to one another. The molecules can then change from a gaseous state to a liquid state since they can no longer maintain their gaseous state
Hence, the reason is not an explanation of the assertion
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Element A has two isotopes. The first isotope is present 19.52% of the time and has a mass of 265.1.
The second isotope has a mass of 182.27. Calculate the atomic mass of element A. (To two decimals
places)
The atomic mass of element A is 198.39 (to two decimal places).
To calculate the atomic mass of element A, we need to consider the relative abundance of each isotope and its corresponding mass. The atomic mass is the weighted average of the masses of all the isotopes, taking into account their relative abundance.
Given:
Isotope 1: Relative abundance = 19.52%, Mass = 265.1
Isotope 2: Relative abundance = 100% - 19.52% = 80.48%, Mass = 182.27
To calculate the atomic mass, we multiply the relative abundance of each isotope by its mass and sum up the results.
Atomic mass = (Relative abundance of Isotope 1 * Mass of Isotope 1) + (Relative abundance of Isotope 2 * Mass of Isotope 2)
Atomic mass = (19.52/100 * 265.1) + (80.48/100 * 182.27)
Calculating the values:
Atomic mass = (0.1952 * 265.1) + (0.8048 * 182.27)
Atomic mass = 51.72752 + 146.661296
Atomic mass = 198.388816
Rounding to two decimal places, the atomic mass of element A is 198.39.
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For each of these pairs of half-reactions, write the balanced equation for the overall cell reaction and calculate the standard cell potential. Express the reaction using cell notation. You may wish to refer to Chapter 20 to review writing and balancing redox equations.
1.
Pt2+(aq)+2e-Pt(s)
Sn2+(aq)+2e-Sn(s)
2.
Co2+(aq)+2e-Co(s)
Cr3+(aq)+3e-Cr (s)
3.
Hg2+(aq)+2e-Hg (I)
Cr2+(aq)+2e-Cr (s)
please help out
1. For the pair of half-reactions:
Pt2+(aq) + 2e- → Pt(s) ... (1)
Sn2+(aq) + 2e- → Sn(s) ... (2)
To obtain the balanced equation for the overall cell reaction, we need to multiply the half-reactions by appropriate coefficients to ensure that the number of electrons transferred is equal. In this case, we can multiply equation (1) by 2 and equation (2) by 1:
2(Pt2+(aq) + 2e-) → 2(Pt(s))
Sn2+(aq) + 2e- → Sn(s)
Combining the equations, we have:
2Pt2+(aq) + Sn2+(aq) → 2Pt(s) + Sn(s)
The cell notation for this reaction is:
Pt2+(aq) | Pt(s) || Sn2+(aq) | Sn(s)
To calculate the standard cell potential (E°), we need to know the standard reduction potentials for Pt2+/Pt(s) and Sn2+/Sn(s) half-reactions. Referring to standard reduction potential tables, we find:
E°(Pt2+/Pt(s)) = +1.20 V
E°(Sn2+/Sn(s)) = -0.14 V
The overall cell potential (E°cell) is the difference between the reduction potentials:
E°cell = E°(cathode) - E°(anode) = 0.00 V - (-0.14 V) = +0.14 V
Therefore, the standard cell potential for this reaction is +0.14 V.
2. For the pair of half-reactions:
Co2+(aq) + 2e- → Co(s) ... (3)
Cr3+(aq) + 3e- → Cr(s) ... (4)
To balance the number of electrons transferred, equation (4) can be multiplied by 2:
2(Co2+(aq) + 2e-) → 2(Co(s))
Cr3+(aq) + 3e- → Cr(s)
Combining the equations, we have:
2Co2+(aq) + Cr3+(aq) → 2Co(s) + Cr(s)
The cell notation for this reaction is:
Co2+(aq) | Co(s) || Cr3+(aq) | Cr(s)
To calculate the standard cell potential (E°), we refer to the standard reduction potentials:
E°(Co2+/Co(s)) = -0.28 V
E°(Cr3+/Cr(s)) = -0.74 V
The overall cell potential (E°cell) is the difference between the reduction potentials:
E°cell = E°(cathode) - E°(anode) = -0.74 V - (-0.28 V) = -0.46 V
Therefore, the standard cell potential for this reaction is -0.46 V.
3. For the pair of half-reactions:
Hg2+(aq) + 2e- → Hg (l) ... (5)
Cr2+(aq) + 2e- → Cr(s) ... (6)
The equation for the overall cell reaction can be obtained by multiplying equation (6) by 2:
2(Hg2+(aq) + 2e-) → 2(Hg (l))
Cr2+(aq) + 2e- → Cr(s)
Combining the equations, we have:
2Hg2+(aq) + Cr2+(aq) → 2Hg (l) + Cr(s)
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For any molecule at the surface the resulting surrounding forces equal zero
True
False
????????????????????????????????????????????????????,,
Which layer of the Earth includes the crust? a atmosphere b hydrosphere c asthenosphere d lithosphere
Answer:
D
Explanation:
Hope that helps :)
What is one way humans can preserve biodiversity
Answer:
Reducing the amount of water you use, by having a 5-minute shower or not running the water when washing up the dishes, can help protect vital wetlands. Plant scientists are also working to help conserve by developing crop varieties that use less water.
Explanation:
Answer:
1. Most of the endangered organisms may represent a source of income.
2. The conservation or preservation may conflict with morals.
3. The role of the species or organisms might not be understood.
Explanation:
Biodiversity simply means the amount or number of living organisms that exist in the world.
The practice of protecting these existing living organisms for specific known purposes is regarded as the preservation or conservation of biodiversity.
This process or practice of preserving biodiversity is important to meet most of the human needs, such as generating income, source of food and fuel among others.
Explain what you think controls a material’s porosity
The primary porosity of a sediment or rock consists of the spaces between the grains that make up that material. The more tightly packed the grains are, the lower the porosity.
How many moles of hydrogen
are in 3.06 x 10^-3 g of glycine C2H5NO2
Answer:
2.04x10⁻⁴ mol
Explanation:
First we convert 3.06x10⁻³ grams of glycine into moles of glycine, using its molar mass:
3.06x10⁻³ g ÷ 75 g/mol = 4.08x10⁻⁵ mol C₂H₅NO₂In order to calculate the number of hydrogen moles, we multiply the number of glycine moles by 5, as there are 5 hydrogen moles per glycine mol:
4.08x10⁻⁵ mol C₂H₅NO₂ * 5 = 2.04x10⁻⁴ mol HA mole equals 6.02 x 10^23 . Answer these questions below.
1. How many ants are in 1.25 moles of ants?
2. How many moles of pencils are in 4.92 * 10^26 pencils?
3. How many molecules are in 0.26 moles of molecules?
4. How many moles of molecules are in 3.46 * 10^19 molecules?
5. 5.3 * 10^20 atoms are equal to how many moles of atoms?
6. 0.11 moles of atoms are equal to how many atoms?
Answer:
1. 1.25 mol ants x 6.02*10^23 ants/1 mol ants = 7.53*10^23 ants
2. 4.92*10^26 pencils x 1 mol pencils/6.02*10^23 pencils = 817 mol pencils
3. 0.26 mol molecules x 6.02*10^23 molecules/1 mol molecules = 1.6*10^23 molecules
4. 3.46*10^19 molecules x 1 mol molecules/6.02*10^23 molecules = 5.75*10^-5 mol molecules
5. 5.3*10^20 atoms x 1 mol atoms/6.02*10^23 atoms = 8.8 mol atoms
6. 0.11 mol atoms x 6.02*10^23 atoms/1 mol atoms = 6.6*10^22 atoms
I would suggest looking into "dimensional analysis" for help with this type of material. Dimensional analysis will stick with you all throughout chemistry, so picking it up will be extremely beneficial.
4.00 g of O2 gas are in a sealed, 2.00 L gas canister at 22.0 °C what is the pressure inside this container (in atm)?
Answer:
1.51448 atms
Explanation:
Which of the following combinations of quantum numbers is permissible?Question 23 options: n = 1, l = 2, ml = 0, ms = n = 4, l = 3, ml = 1, ms = n = 3, l = 3, ml = 1, ms = n = 2, l = 1, ml = –1, ms = 0 n = 4, l = 3, ml = 4, ms =
Answer: n= 4, , l= 3, [tex]m_l[/tex] = 1, permissible
Explanation:
Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1).For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as [tex]m_l[/tex]. The value of this quantum number ranges from [tex](-l\text{ to }+l).[/tex]
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]. The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
a) n=1 , l= 2, [tex]m_l[/tex] = 0, not permissible as l can not greater than n.
b) n= 4, , l= 3, [tex]m_l[/tex] = 1, permissible
c) n= 3, l= 3, [tex]m_l[/tex] = 1, not permissible as l can not equal than n.
d) n= 4 , l= 3, [tex]m_l[/tex] = 4, not permissible as [tex]m_l[/tex] can not greater than l.
A change in velocity can occur without a change in speed *
A: True
B: False?
Answer: True
Explanation: For example, changing direction can change velocity
List four reasons why the temperature of a flame is lower than the adiabatic flame temperature."
Answer:
substances is never entirely a one way
Explanation:
no possibility of obtaining complete combustion at hig
temperature.
always loss of heat from the temperature
its temperature is lower than the constant volume process because some of the energy is utilized to change the volume of the system
Four reasons why the temperature of flame is lower than adiabatic flame temperature is that there is no possibility of obtaining complete combustion at high temperature due to high thermal energy.There is always loss of heat from temperature,its temperature is lower than constant volume process because some of energy is utilized to change the volume of the system.
What is thermal energy?Thermal energy is defined as a type of energy which is contained within a system which is responsible for temperature rise.Heat is a type of thermal energy.It is concerned with the first law of thermodynamics.
Thermal energy arises from friction and drag.It includes the internal energy or enthalpy of a body of matter and radiation.It is related to internal energy and heat .It arises when a substance whose molecules or atoms are vibrating faster.
These vibrating molecules and atoms collide and as a result of which heat is generated in a substance , more the collision of particles , higher is the thermal energy.
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A sample of chlorine gas has a volume of 0.30 L at 273 K and 1 atm pressure. What temperature (in ∘C) would be required to increase the volume to 1.0 L ?
Answer:
[tex]T_2=637\°C[/tex]
Explanation:
Hello there!
In this case, considering this problem as pressure constant, since the change is exhibited in temperature and volume only, it is possible for us to use the Charles' law as shown below:
[tex]\frac{V_2}{T_2}=\frac{V_1}{T_1}[/tex]
Thus, by solving for the final temperature, we obtain:
[tex]T_2=\frac{T_1V_2}{V_1}\\\\T_2=\frac{273K*1L}{0.30L}\\\\T_2=910K-273\\\\T_2=637\°C[/tex]
Best regards!
Mixtures that would be considered buffers include which of the following? I. 0.10 M HCl + 0.10 M NaCl II. 0.10 M HF + 0.10 M NaF III. 0.10 M HBr + 0.10 M NaBr
Answer:
II. 0.10 M HF + 0.10 M NaF.
Explanation:
A buffer system may be formed in one of two ways:
A weak acid and its conjugate base.A weak base and its conjugate acid.Which of the following mixtures could be considered a buffer?
I. 0.10 M HCl + 0.10 M NaCl. No, because HCl is a strong acid.
II. 0.10 M HF + 0.10 M NaF. Yes, since HF is a weak acid and F⁻ (coming from NaF) is its conjugate base.
III. 0.10 M HBr + 0.10 M NaBr. No, because HBr is a strong acid.
are two double bonds in
uble dash.
:0=C=O
NO2-
Answer:
Yes
Explanation:
There are in double dash :) ( = )
Answer:
yes
Explanation:
Because the double dash shows a double bond it simply means that two electrons from each atom are being shared
calorimetry and specific heat lab report can someone please write this for me
If anyone still needed help on this, someone wrote out the student table on brainly.com, look up Lab: Calorimetry and Specific Heat.
The page should say something along the lines of please Please i need it Lab: Calorimetry and Specific Heat
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Answer:
If anyone still needed help on this, someone wrote out the student table on brainly.com, look up Lab: Calorimetry and Specific Heat.
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Explanation:
125.0 mL of 2.00 M NaCl solution is diluted to a concentration of 1.50 M. How many mL of water was added to the original volume? (Hint: must find V2 first) 1
166.7 ml
30.9 mL
41.7 ml
292 mL
(no links and please show work)
Answer:
Option A
Explanation:
As we know
C1V1 = C2V2
C1 = concentration of solution before dilution
V1 = Volume of solution before dilution
C2 = concentration of solution after dilution
V2 = Volume of solution after dilution
Substituting the given values in above equation, we get -
125 mL * 2.00 M = X mL * 1.50 M
X mL = 125 mL * 2.00 M / 1.50 M
X = 167 mL
Hence, option A is correct
how to calculate relative abundance of copper isotopes
Answer: Atomic mass (Cu) = (x)(63.0 amu) + (1 – x)(65.0 amu) = 63.5 amu 63.0x + 65 – 65.0x = 63.5 –2x = –1.5 x = 0.75 The percent abundance of each isotope is 75.0 % (Cu-63) and 25.0 % (Cu-65).
Explanation: As you know, the average atomic mass of an element is determined by taking the weighted average of the atomic masses of its naturally occurring isotopes.
Simply put, an element's naturally occurring isotopes will contribute to the average atomic mass of the element proportionally to their abundance.
Question 10 of 23
Which characteristic makes a digital signal more useful than an analog signal
for storing information?
A. A digital signal includes a spectrum of colors.
B. A digital signal is more difficult to copy.
C. A digital signal cannot be heard by humans.
O D. A digital signal moves between a discrete number of values.
Answer: O D. A digital signal moves between a discrete number of values.
Explanation:
The digital signals are considered more reliable over the analog signals as they encode the information in a coded form. The bits or samples of the data are transmitted and converted into digital and numerical value. The stream of encoded data is in the form of continuous data at regular time intervals. It provides information in waveform and the data is in compact form. The data is in the form of binary bits 0 and 1 so greater the number of bits greater will be the greater will be the resolution of the information.