Answer:
from the words below underline six example of rhetorical patterns
A chemical reaction involves the reactants A and B, and the product C.
A+B→C
Trial 1 2 3
[A](mol/L) 0.10 0.10 0.20
[B](mol/L) 0.10 0.20 0.10
−Δ[A]Δt(molL s) 3.08×10−9 2.464×10−8 1.232×10−8
Explanation:
a+b =ab
answer
ab added all
is give answer
2 nitrogen atoms and five chlorine atoms what compound does that make
Answer:
dinitrogen pentachloride
If 5 g of sodium chloride saturates 12.5 g of water at 10 °C, what mass of sodium chloride would saturate 50 g of water at constant temperature?
Explanation:
since 5g saturates 12.5g of water at 10°c
so......x would saturate 50g of water at 0°c
then you can cross multiply
The anion O3− does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Indicate the values of nonzero formal charges and include lone pair electrons.
One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.
Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.
We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.
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Why is the first one (A) correct?
Answer: yes it is correct
Explanation: the higher it is the cooler.
1 year = 365 days
1 day = 24 hours
How many years is 1,000,000 hours?
Has to be dimensional analysis
what is the expected hybridization of the central atom tetrahedral
Answer:
Tetrahedral molecules are normally spy hybridized.
Explanation:
) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at 25 °C, and the density of water at 25 °C is 0.9970 g>mL)
At the left, grains of the mineral calcite float on the surface of the liquid bromoform (d = 2.890 g/mL) At the right, the grains sink to the bottom of liquid Chloroform (d = 1.444 g/mL). By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined
Hey there!
It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.
In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:
[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]
Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:
[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]
Thus, the density of the calcite sample will be:
[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]
This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL
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Draw the other possible resonance structure of each organic ion. In each case, draw the structure that minimizes formal charges. Be sure to include all appropriate nonbonding electrons and charges. A three carbon chain with a double bond between carbons 1 and 2. Carbon 1 is bonded to two hydrogen atoms, carbon two is bonded to one hydrogen atom, and carbon three is bonded to two hydrogen atoms. There is a plus one charge on the third carbon atom. Draw the resonance structure of the allyl ion. A carbon atom is single bonded to a C H 3 group, double bonded to an oxygen atom, and single bonded to an N atom. The N atom is bonded to a hydrogen atom. It has two lone pairs and a minus one charge. The O atom has two lone pairs. Draw the resonance structure of the amidate ion.
Three resonance structures can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.
Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.
The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.
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Determine the mass in grams of 5.15 × 10²¹ atoms of chromium. (The mass of one mole of chromium is 52.00 g.)
The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
From Avogadro's hypothesis,
6.02×10²³ atoms = 1 mole of Cr
But:
1 mole of Cr = 52 g
Thus, we can say that:
6.02×10²³ atoms = 52 g of Cr
With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:
6.02×10²³ atoms = 52 g of Cr
Therefore,
5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³
5.15×10²¹ atoms = 0.44 g of Cr
Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
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A mixture of 3.25 moles of oxygen gas and 2.75 moles of nitrogen gas exert a total pressure of 22.4 atm. What is the partial pressure of oxygen
The partial pressure of oxygen is equal to 10.2 atm.
Given the following data:
Number of moles of oxygen = 3.25 moles.Number of moles of nitrogen = 2.75 moles.Total pressure = 22.4 atm.To determine the partial pressure of oxygen:
First of all, we would find the total number of moles of the elements:
[tex]n=3.25+2.75[/tex]
n = 6 moles
Next, we would determine the mole fraction of the oxygen by using this formula:
[tex]Molefraction \;of \;a \;substance =\frac{No.\; of \; moles \;of \;substance}{Total \;no. \;of \; moles \;of \;substances}[/tex]
Substituting the values, we have:
[tex]Molefraction \;of \;a \;substance =\frac{2.75}{6} \\\\Molefraction \;of \;a \;substance =0.4583[/tex]
For oxygen:
[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.4583 \times 22.4[/tex]
Partial pressure of oxygen = 10.2 atm.
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Answer:
12.1
Explanation:
3.25 + 2.75= 6 moles total
then take: 3.25/6 to get 0.542
multiply 0.542 by 22.4 to get 12.1
the other answer solved for nitrogen, not oxygen.
The mass of a metal ball was 20.350 g. The volume of the water in a 25 mL graduated cylinder was increased from 16.05 to 17.90 mL when the metal ball was placed in the cylinder. What is the density of the metal ball
Answer:
11000 g/L
Explanation:
Since density is mass divided by volume, find the volume of the metal ball by subtracting the initial volume of water in the graduated cylinder from the ending volume. In this case, the volume would be 17.90 mL - 16.05 mL = 1.85 mL = 0.00185 L. Then using the formula for density, calculate 20.350 g / 0.00185 L = 11000 g/L.
Density is an intensive property as it does not depend on the quantity of the substances Whereas mass and volume are extensive property. Therefore, the density of the metal ball is 11000 g/L.
What is density?Density tells about the compactness of the substances, how much dense is the substances in other words. Object that is more denser than water they just sink in the water.
Mathematically,
Density = Mass of the metal ball ÷volume marked on the graduated cylinder
volume=ending volume- initial volume of water in the graduated cylinder
=17.90 mL - 16.05 mL
= 1.85 mL
= 0.00185 L.
substituting all the given values in the above equation, we get
density= 20.350 g / 0.00185 L
= 11000 g/L.
Therefore, the density of the metal ball is 11000 g/L.
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HELP ME OUT PLEASE!!!!!!!!!
Compare and contrast model A with model C. How are they alike: How are they different?
A) They are composed of different elements in the same ratio.
B) They are composed of the same elements in different ratios.
C) They are composed of different elements in different ratios
D) The are composed of the same elements but the bond types differ
Answer:
Model C has more double the oxygen than A. I think it is B
Explanation:
2. Alex pulls on the handle of a claw hammer with a force of 15 N. If
the hammer has a mechanical advantage of 5.2, how much force
is exerted on the nail in the claw?
Answer:
78n
Explanation:
The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.
What is the mechanical advantage?The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.
The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.
Given, the input force = 15 N
The mechanical advantage of the hammer = 5.2
Mechanical advantage = Output force/ Input force
5.2 = Output/15
Output force = 15 ×5.2 = 78 N
Therefore, the force is exerted on the nail in the claw is equal to 78 N.
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The sum of the number of proteins and neutrons in an atoms nucleus is its __________ ___________.
Answer:
Mass Number
Explanation:
In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.
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a material that is not a mixture; has the same properties all the way through
Answer:
Explanation:
The material that is not a mixture; it has the same properties all the way through is called a substance. Thus the material that is not a mixture; it has the same properties all the way through is called a substance.
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pls help 8th grade k12
________ are neutral particles found in the nucleus of an atom.
Compound A is neutral and Compound B is acidic. Both are water-insoluble solids. A and B are dissolved in dichloromethane (DCM) and extracted with aqueous base. The layers are then separated. What must be done to obtain the compound in the aqueous layer
In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.
When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
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A sample of crude sodium iodide was analysed by the following rection.
I- + SO42- → I2 + H2S (unbalanced)
a)Determine the oxidation number for S in SO42-.
b)The above reaction requires a basic medium. Write the balanced chemical equation. [6 marks]
A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.
8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
Let's consider the following unbalanced redox reaction.
I⁻ + SO₄²⁻ → I₂ + H₂S
The oxidation number of I goes from -1 (I⁻) to 0 (I₂) so it is oxidized.The oxidation number of S goes from +6 (SO₄²⁻) to -2 (H₂S) so it is reduced.The corresponding half-reactions are:
I⁻ → I₂
SO₄²⁻ → H₂S
We will perform the mass balance adding OH⁻ and H₂O where appropriate.
2 I⁻ → I₂
6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻
Then, we will perform the charge balance adding electrons where appropriate.
2 I⁻ → I₂ + 2 e⁻
8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻
Finally, we will multiply the first half-reaction by 4 and the second by 1, and add them.
4 × (2 I⁻ → I₂ + 2 e⁻)
1 × (8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻)
------------------------------------------------------------
8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.
8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
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Is brass a solution:
Answer:
Brass is an alloy, and either a "solid solution".
Alloys in general may be solid solutions or they simply be mixtures
Explanation:
Hope it Helps you!!The table shows the order of the Sun and some of the planets in our solar system.
The letters represent the Earth, Neptune and the Sun, so match them up in your answer.
Explanation:
A=Sun
B=Earth
C=Neptune.
hope my answer will definitely help you out.
The box shows the chemical formula for bleach.
Yes thank you Great explanation
Answer:
you're welcome
Chemistry
Help me please.
Answer:
4:6
Explanation:
Which of the following best describes the scientific exploration of the atom?
The result of a division problem is the a ) divisor . b ) quotient . c ) factor . d ) remainder .
Answer:
The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.
Explanation:
GIVING BRAINLY AND 20 POINTS
A sound wave in air has the wavelength of 1.36 m. Calculate its frequency? Assume the speed is 340 m/s.
Answer: the answer is 0.074Hz
Explanation:
Given, (In air)
Velocity V=340m/s
Frequency f=20,000Hz
Wavelegth λ=?
V=f.λ
λ=
F
V
=
20,000
340
=0.017Hz
Also, Given (in Water)
Velocity, V=1480m/s
Frequencyf=20,000Hz
wavelength, λ=?
V=F.λ
λ=
F
V
=
20,000
1480
=0.074Hz
Answer:
frequency
Explanation:
frequency is velocity/ wavelength
340/1.36
250
Escriba un poema usando 4 de las siguientes palabras: lluvia, nieve, lluvia, granizo, tornado, huracanes, frente cálido, frente frío, alta presión, baja presión, termómetro, barómetro, anemómetro, pluviómetro, veleta o higrómetro.
plis ayúdenme porfa
Answer:
i dont speak mexican
Explanation:
. Nhiệt độ ban đầu của 344 g một mẫu sắt là 18,2oC. Nếu mẫu sắt này hấp thụ 2,25 kJ nhiệt lượng thì nhiệt độ cuối của mẫu sắt này là bao nhiêu? (CFe = 0,449 J/g.K).
Answer:
i can't understand the question