How far will you travel in 180 seconds running at a rate of 6 m/s?
Answer:
1080 meters.
Explanation:
A liquid has a volume of 4mL and a mass of 24 grams. What is the density of the liquid?
Answer:
the density is 6
Explanation:
mass divided by volume equals density
A solution is made by dissolving
42.3 g of potassium hydroxide
(KOH) in 329 g of water.
What is the molality of the solution?
Answer:
2.29 m.
Explanation:
The following data were obtained from the question:
Mass of KOH = 42.3 g
Molar mass of KOH = 56.11 g/mol
Mass of water = 329 g
Molality of KOH = ?
Next, we shall determine the number of mole in 42.3 g of KOH. This can be obtained as follow:
Mass of KOH = 42.3 g
Molar mass of KOH = 56.11 g/mol
Mole of KOH =?
Mole = mass /Molar mass
Mole of KOH = 42.3/56.11
Mole of KOH = 0.754 mole
Next, we shall convert 329 g of water to kilogram (kg). This can be obtained as follow:
1000 g = 1 kg
Therefore,
329 g = 329 g /1000 g × 1 kg
329 g = 0.329 kg
Therefore, 329 g of water is equivalent to 0.329 kg
Finally, we shall determine the molality of the KOH solution ad follow:
Molality is defined as the mole of solute per unit kilogram of solvent (water) i.e
Molality = mole/ mass (kg) of water
Mole of KOH = 0.754 mole.
Mass of water = 0.329 kg.
Molality = mole/ mass (kg) of water
Molality = 0.754/0.329
Molality = 2.29 m
Therefore, the molality of the KOH solution is 2.29 m.
Answer:
2.29
Explanation:
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an actively growing forest. The atmosphere is approximately 20% O2 and 80% N2, but contains 0.039% CO2 by weight.
A) How much carbon is present in the entire atmosphere lying above each square meter of the earth's surface?
B) At the current rate of utilization, how long would it take to use all the CO2
in the entire atmosphere directly above a forest?
Answer:
a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg
b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years
Explanation:
first we calculate the moles of carbon
moles = mass/molar mass
= 1kg/12gmol⁻¹
= 1000g/12gmol⁻¹
= 83.33 mol
now using the ideal gas equation
we find the volume of co₂required based on 83.33 moles
PVco₂ = nRT
Vco₂ = nRT/P
Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm
Vco₂ = 2083.73 L
so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon
therefore
Vair × 0.0390/100 = 2038.73L
Vair = (2038.73L × 100) / 0.0390
Vair = 5.23 × 10⁶L
therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon
a)
Here we calculate the mass of air over 1 square meter of surface.
Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²
NOW
mass of air = 1.020×10⁴kgm⁻² × 1m²
= 1.020×10⁴kg
= 1.020×10⁷g [1kg = 10³g]
we now find the moles of air associated with it
moles = mass/molar mass
= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)
= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)
= 1.020 × 10⁷g / 28.8 gmol⁻¹
= 354166.67mol
so based on the question, for each mole (air), there is 0.0390% of CO₂
now to calculate the moles of CO₂ we say;
MolesCo₂ = 0.0390/100 × 354166.67mol
= 138.125 moles
Now we calculate mass of CO₂ from the above findings
Moles = mass/molar mass
mass = moles × molar mass
= 138.125 moles × 12gmol⁻¹
= 1657.5g
we covert to KG
= 1657.5g / 1000
mass = 1.65kg
therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg
b)
to find the number years required to use up all the CO₂, WE SAY
Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year
Number of years = 1.65kgm⁻² / 1kg²year⁻¹
Number of years = 1.65 years
Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years
Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.
Answer: 2.7 cm
Explanation: ~~~~~ there ya go
Question 13 of 16
The mass of a piece of metal is 25.4253 grams. When the piece of
metal was dropped into a graduated cylinder that had an initial water
volume of 6.85 cm, the total volume was 9.84 cm? What is the
density of the piece of metal in g/cm??
Answer:
[tex]\rho =8.50g/cm^3[/tex]
Explanation:
Hello,
In this case, due to the volume difference caused by the addition of the metal, one could notice that the volume of the metal is:
[tex]V_{metal}=9.84cm^3-6.85cm^3=2.99cm^3[/tex]
In such a way, given the mathematical definition of density, it turns out:
[tex]\rho =\frac{m}{V}=\frac{25.4253g}{2.99cm^3}\\\\\rho =8.50g/cm^3[/tex]
Regards.
When 1 carbon atom combines with 2 oxygen atoms, the resulting substance is called a
Answer:
Carbon Dioxide
Explanation:
Which is a compound:)
I need these questions answered
Answer:
Three objects with kinetic energy
A ball rolling down the street
Moving Car
Bullet
Law of Conservation of Energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
Though technically there are limitless forms of both types of energy, officially there's five types of kinetic energy: radiant, thermal, sound, electrical and mechanical and potential energy adds gravitational, nuclear, and elastic.
If 6.09 g of sodium chloride are mixed with 35.50 g of water, what is the mass % (w/w%) salt in the solution
Answer:
14.6 %
Explanation:
Step 1: Given data
Mass of sodium chloride (solute): 6.09 g
Mass of water (solvent): 35.50 g
Step 2: Calculate the mass of solution
We will use the following expression.
m(solution) = m(solute) + m(solvent)
m(solution) = 6.09 g + 35.50 g
m(solution) = 41.59 g
Step 3: Calculate the percent by mass of the salt
We will use the following expression.
%w/w = mass of NaCl / mass of solution × 100%
%w/w = 6.09 g / 41.59 g × 100%
%w/w = 14.6 %
At a certain temperature this reaction follows first-order Kinetics with a rate constant of 0.0660
2H1 (g)----------> H2, (g)+I2
Suppose a vessel contains HI at a concentration of 0.310 M. Calculate how long it takes for the concentration of HI to decrease to 0.0558 M. You may assume no other reaction is important. Round your answer to 2 significant digits.
Answer:
After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M.
Explanation:
Based on the reaction of the problem, you have as general kinetic law for a first-order reaction:
ln[HI] = -kt + ln [HI]₀
Where [HI] is actual concentration after time t,
k is rate constant
and [HI]₀ is initial concentration of the reactant.
Initial concentration of HI is 0.310M,
K is 0.0660s⁻¹,
And the actual concentration is 0.0558M:
ln[HI] = -kt + ln [HI]₀
ln[0.0558M] = -0.0660s⁻¹*t + ln [ 0.310M]
-1.7148 = -0.0660s⁻¹*t
26.0s = t
After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M
For the following amino acid, the name, three-letter abbreviation, or one-letter abbreviation is given. Complete the missing information.Name: proline Three letter abbreviation:____________One letter abbreviation:________
Answer:
The correct answer is "Pro; P".
Explanation:
Amino acids are not only identified by its full name, there are three-letter and one-letter abbreviations for each amino acid that helps to annotate sequences and biological processes. In the case of proline, the three letter abbreviation is "Pro" and the one letter abbreviation is "P". Proline got its name for its cyclic structure, that resembles the structure of pyrrole.
3y= 4y^3
Write a verbal sentence to represent each equation.
Answer:
See explanation
Explanation:
To write a verbal sentence to represent an equation, we must look at the equation closely in order to capture its essence.
For this equation, 3y= 4y^3, we can write it in words as:
Three multiplied by y is equal to four multiplied by y raised to power three.
Use the observation in the first column
Observation Questions
At 4 °C, Substance E has a Which has a higher boiling point
vapor pressure of 86. torr and Substance A. Substance E
F has a vapor pressure of 136. torr. B. Substance F
C. Neither, E and F have the
same boiling point.
The enthalpy of vaporization of Substance At any temperature where both
C is smaller than that of Substance D. substances are liquid which has
has the higher vapor pressure?
A. Substance C
B. Substance D
C. Neither, C and D have the
same vapor pressure
Which has a higher vapor pressure?
At 1 atm pressure Substance A boils at A. Substance A
129. °C and Substance B B. Substance B
boils at 154. °C. C. Neither, A and B have the
same vapor pressure.
Answer:
A. Substance E
A. Substance C
A. Substance A
Explanation:
Given that:
At 4 °C, Substance E has a vapor pressure of 86. torr and Substance F has a vapor pressure of 136. torr
Which has a higher boiling point?
A. Substance E
B. Substance F
C. Neither,EandF have the same boiling point
The vapor pressure varies inversely proportional to the boiling point.
[tex]\mathbf{vapor \ pressure \ \ \alpha \ \ \dfrac{1}{boiling \ point}}[/tex]
Therefore, the lower the vapor pressure, the higher the boiling point.
At 4°C, Substance E with a lower vapor pressure of 86. torr will have a higher boiling point from the given information.
2.
Recall that :
[tex]\mathbf{vapor \ pressure \ \ \alpha \ \ \dfrac{1}{enthalpy \ of \ vaporization}}[/tex]
therefore, the lower the enthalpy of vaporization, the higher the vapor pressure at any given temperature.
Given that:
Substance C has an enthalpy of vaporization smaller than that of substance D. Then, substance C has a higher vapor pressure.
3.
We've earlier said that:
The vapor pressure varies inversely proportional to the boiling point.
[tex]\mathbf{vapor \ pressure \ \ \alpha \ \ \dfrac{1}{boiling \ point}}[/tex]
Therefore, the lower the vapor pressure, the higher the boiling point.
As such, Substance A will have a higher boiling point.
What is an extensive property? *
A property that changes if temperature changes
A property that will NOT change if temperature changes
A property that changes if the amount of substance changes
A property that does NOT change if the amount of substance changes
Help :( pls
Answer:
A property that changes if the amount of substance changes
Explanation:
An extensive property is a property that depends on the amount of matter in a sample.
A runner competed in a 5-mile run. How many yards did she run?
a-8800 yards
b-8800 miles
с-8657 yards
What is a factual statement about how nature behaves or functions?
a hypothesis
O a law
O a theory
an experiment
Answer:its a law
Explanation:
I passed that quiz 100%, my answer for that question was law
so its law
An atom's Lewis dot structure has four dots. Which of the following elements could it be, and why?
Answer: Carbon, because it is in group 14 and has four valence electrons
Explanation: Just did this quiz
Answer:carbon
Explanation:
In reading a line drawing, how do you know where atoms of these elements are in the structure if they are missing from the drawing?
Answer:
The atoms of an element are represented in a chemical line drawing with its chemical formula.
Explanation:
chemical structural drawing helps to represent the pattern for which an element is formed. Chemical elements are made up of atoms that represent their single state.
The line drawing is made up of lines (representing the chemical bond between atoms) and the atoms or various atoms that make up the element.
A buffer contains 0.010 moles of lactic acid (pKa = 3.86) and 0.050 mol of sodium lactate per liter.
(a) Calculate the pH of the buffer.
(b) Calculate the change in pH when 5 mL of 0.5 M HCl is added to 1 L of the buffer.
(c) What pH change would you expect if you added the same quantity (5 mL) of HCl to 1 L of pure water?
Answer:
a. pH = 4.56
b. Change in pH = 0.85
c. Change in pH = 5.4
Explanation:
a. The pKa of lactic buffer is: 3.86.
Using Henderson-Hasselbalch formula for the lactic buffer:
pH = 3.86 + log [Lactate] / [Lactic acid]
Where [] is molarity of each compound but could be taken as moles
Replacing:
pH = 3.86 + log [0.050 moles] / [0.010 moles]
pH = 4.56
b. The HCl added reacts with Lactate producing lactic acid. Moles of HCl are:
5x10⁻³L * (0.5mol /L) = 0.025 moles HCl
Moles of lactate: 0.050moles - 0.025 moles = 0.025 moles
Moles lactic acid: 0.010 moles + 0.025 moles = 0.035 moles
pH = 3.86 + log [0.025 moles] / [0.035 moles]
pH = 3.71
Change in pH = 4.56 - 3.71 = 0.85
c. 1L of pure water has a pH of 7. 0.025 moles of HCl = 0.025 moles H⁺ in 1.005L:
0.025 mol / 1.005L = 0.0249M = [H⁺]
As pH = -log [H⁺]
pH = 1.6
Change in pH = 7.0 - 1.6 = 5.4
Rank the following atoms in order of increasing electronegativity.a. Se, O, S b. P, Na, Cl c. Cl, S, F d. O, P, N
Answer:
a)- Se, S, O
b)- Na, P, Cl
c)- S, Cl, F
d)- P, N, O
Explanation:
We can solve this problem by looking at the Periodic Table. In a group, electronegativity increases from bottom to top; whereas in periods it increases from left to right.
Thus, we order the elements from lower to higher electronegativity as follows:
a. Se, O, S ⇒ order: Se, S, O
Because they are all in the same group. Se is near the bottom, followed by S and O is at the top.
b. P, Na, Cl ⇒ order: Na, P, Cl
They are in the same period. Na is at the left, followed by P and Cl is nearest the right.
c. Cl, S, F ⇒ order: S, Cl, F
P and Cl are in the same period, and P is at the left, so it has the lowest electronegativity. F is in the same group of Cl, but at the top. F has the highest electronegativity.
d. O, P, N ⇒ order: P, N, O
N and P are in the same group, but P is at the bottom so it has the lower electronegativity. N and O are in the same period, but O is at the right, so it is the most electronegative.
Sort the following analytical techniques as either classical methods or instrumental methods.
a) surface analysis
b) precipitation titration
c) gravimetric analysis
d) high performance liquid chromatography
e) potentiometry
f) atmoic spectroscopy
Answer:
Instrumental methods
surface analysis
high performance liquid chromatography
atomic spectroscopy
potentiometry
Classical methods
precipitation titration
gravimetric analysis
Explanation:
Instrumental methods of analysis are those analytical methods in which the responsibility of detection has been removed from human beings and placed on automated instruments while classical methods are those analytical methods in which the responsibility of detection remains the responsibility of human beings.
Many instrumental methods such as HPLC rely on computer screens as readout devices.
What is the correct answer with the appropriate significant digits?6.25 x 19.50
Which tool can be used to measure the volume of liquid?
an electronic balance
a meniscus
a caliper
a beaker
Answer:
The answer is D a beaker
Explanation:
Hope this helps :D
Which of the following DOES NOT have 2 significant
figures?
11,000,000,000
0.11
1.001
0.0000011
Answer:
1.001
Explanation:
The Significant Figures are 1 0 0 1, This answer has 4 Significant figures, while the other three have only 2 significant figures
The enthalpy of sublimation of iodine is 60.2 kJ/mol, and its enthalpy of vaporization is 45.5 lz.1/mol. What is the enthalpy of fusion of iodine?
Answer:
ΔH = 14,7kJ/mol
Explanation:
It is possible to make algebraic sum of several chemical process to obtain enthalpy of a determined reaction (Hess's law).
Sublimation of iodine is (Transition from solid to gas):
I₂(s) → I₂(g) ΔH = 60.2kJ/mol
Vaporization of iodine (From liquid to gas):
I₂(l) → I₂(g) ΔH = 45.5kJ/mol
Fusion of iodine (From solid to liquid can be obtained subtracting the sublimation process - Vaporization process:
I₂(s) → I₂(g) ΔH = 60.2kJ/mol
I₂(g) → I₂(l) ΔH = -45.5kJ/mol
I₂(s) → I₂(l) ΔH = 60.2kJ/mol - 45.5kJ/mol
ΔH = 14,7kJ/molDescribe how matter is classified into mixtures, pure substances, elements, and compounds.
Answer:
Matter can be broken down into two categories: pure substances and mixtures.
Pure substances are further broken down into elements and compounds. ... A chemical substance is composed of one type of atom or molecule.
A mixture is composed of different types of atoms or molecules that are not chemically bonded.
hope this helped
Matter is divided in two categories: pure substances and mixtures.Pure substances are further broken down into elements and compounds.Chemical substance is composed of one type of atom or molecule.A mixture is composed of different types of atoms or molecules .
What is matter?Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.
Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.
The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.
Learn more about matter,here:
https://brainly.com/question/12972782
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Disadvantage of dot structure
How many atoms of titanium are there in 0.125 mole of each of the following?a. ilmenite, FeTiO3b. titanium(IV) chloridec. Ti2O3d. Ti3O5
Answer:
a. 7.528x10²² atoms Ti
b. 7.528x10²² atoms Ti
c. 1.5055x10²³ atoms Ti
d. 2.258x10²³ atoms Ti
Explanation:
Using the molecular formula of the structures, we can determine the moles of titanium in 0.125 moles of each compound. With moles of titanium we can convert these to atoms using Avogadro's number as follows:
a. FeTiO₃: 1 mole of Ti per mole of FeTiO₃.
Moles of Ti are 0.125 moles. Atoms are:
0.125 moles Ti * (6.022x10²³ atoms / 1 mole) =
7.528x10²² atoms Tib. Titanium (IV) chloride = TiCl₄: 1 mole of Ti per mole of TiCl₄.
Moles of Ti are 0.125 moles. Atoms are:
0.125 moles Ti * (6.022x10²³ atoms / 1 mole) =
7.528x10²² atoms Tic. Ti₂O₃: 2 moles of Ti per mole of Ti₂O₃.
Moles of Ti are 2*0.125 moles = 0.25 moles. Atoms are:
0.25 moles Ti * (6.022x10²³ atoms / 1 mole) =
1.5055x10²³ atoms Tia. Ti₃O₅: 3 mole of Ti per mole of Ti₃O₅.
Moles of Ti are 3*0.125 moles = 0.375 moles of Ti. Atoms are:
0.375 moles Ti * (6.022x10²³ atoms / 1 mole) =
2.258x10²³ atoms TiFrom the formula for calculating number of atoms, the number of atoms of Ti in 0.125 moles of the compounds FeTiO₃, TiCl₄, Ti₂O₃, Ti₃O₅ are 7.528 * 10²² atoms Ti, 7.528 x 10²² atoms Ti, 1.5055 * 10²³ atoms Ti and 2.258 * 10²³ atoms Ti respectively.
What is the number of atoms present in the compounds?The number of atoms present in a substance is calculated using the formula:
Number of atoms = Number of moles * 6.022 * 10³Moles of Titanium in the compounds:
a. FeTiO₃ contains 1 mole of Ti per mole of FeTiO₃.
Moles of Ti are 0.125 moles.
number of atoms = 0.125 * 6.022x10²³ atoms
number of atoms = 7.528 * 10²² atoms Ti
b. 1 mole of Titanium (IV) chloride TiCl₄ contains 1 mole of Ti per mole of TiCl₄.
Moles of Ti are 0.125 moles.
Atoms of titanium = 0.125 * 6.022x10²³ atoms / 1 mole)
number of atoms = 7.528 x 10²² atoms Ti
c. 1 mole of Ti₂O₃ contains 2 moles of Ti per mole of Ti₂O₃.
Moles of Ti are 2 * 0.125 moles = 0.25 moles
number of atoms = 0.25 * 6.022x10²³ atoms
number of atoms = 1.5055 * 10²³ atoms Ti
d.1 mole of Ti₃O₅ contains 3 mole of Ti per mole of Ti₃O₅.
Moles of Ti are 3 * 0.125 moles = 0.375 moles of Ti.
number of atoms = 0.375 * 6.022x10²³ atoms
number of atoms = 2.258 * 10²³ atoms Ti
Therefore, the number of atoms of Ti in 0.125 moles of the compounds FeTiO₃, TiCl₄, Ti₂O₃, Ti₃O₅ are 7.528 * 10²² atoms Ti, 7.528 x 10²² atoms Ti, 1.5055 * 10²³ atoms Ti and 2.258 * 10²³ atoms Ti respectively.
Learn more about number of atoms at: https://brainly.com/question/15193101
What sequence should be followed when conducting a laboratory investigation? Make observations, gather experimental data, form a conclusion, state a problem Define a problem, form a hypothesis, gather experimental data, form a conclusion. Form a hypothesis, form a conclusion, gather experimental data, define a problem Gather experimental data, make observations, form a conclusion, for a hypothesis.
Answer:
Define a problem, form a hypothesis, gather experimental data, form a conclusion
Answer:
Make observations, gather experimental data, form a conclusion, state a problem Define a problem, form a hypothesis, gather experimental data, form a conclusion.
Explanation:
It takes 330 joules of energy to raise the temperature of 24.6 gbenzene from 21 degrees Celsius to 28.7 degrees Celsius at constantpressure. What is the molar hear capacity of benzene at constantpressure?
Given :
Energy , E = 330 J .
Initial temperature , [tex]T_i=21^oC[/tex] .
Final temperature , [tex]T_f=24.6^oC[/tex] .
Mass of benzene , m = 24.6 g .
To Find :
The molar hear capacity of benzene at constant pressure .
Solution :
Molecular mass of benzene , M = 78 g/mol .
Number of moles of benzene :
[tex]n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol[/tex]
Energy required is given by :
[tex]q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}[/tex]
Hence , this is the required solution .