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2 years ago

Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample

Chemistry

1 answer:

Dimas [21]2 years ago

3 0

Answer:

3.27·10²³ atoms of O

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05 $\frac{g}{mol}$ .

We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.

19.3g Na₂SO₄ · $\frac{1 mol Na^2SO^4}{142.05g Na^2SO^4}$ · $\frac{4 mol O}{1 mol Na^2SO^4}$ · $\frac{6.022x10^2^3}{1 mol O}$

After doing the math for this dimensional analysis, you should get a quantity of approximately 3.27·10²³ atoms of O.

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

2 years ago

An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom

diamong [38]

Answer:

Explanation:

answer is a on edg

6 0

2 years ago

A medical lab is testing a new anticancer drug on cancer cells. The drug stock solution concentration is 1.5×10−9m, and 1.00 ml

Elis [28]

The concentration of the drug stock solution is 1.5*10^-9 M i.e. 1.5 * 10^-9 moles of the drug per Liter of the solution

Therefore, the number of moles present in 1 ml i.e. 1*10^-3 L of the solution would be = 1 *10^-3 L * 1.5 * 10^-9 moles/1 L = 1.5 * 10^-12 moles

1 mole of the drug will contain 6.023*10^23 drug molecules

Therefore, 1.5*10^-12 moles of the drug will correspond to :

1.5 * 10^-12 moles * 6.023*10^23 molecules/1 mole = 9.035 * 10^11 molecules

The number of cancer cells = 2.0 * 10^5

Hence the ratio = drug molecules/cancer cells

= 9.035 *10^11/2.0 *10^5

= 4.5 * 10^6

8 0

2 years ago

Read 2 more answers

Water treatment plants commonly use chlorination to destroy bacteria. a byproduct is chloroform (chcl3), a suspected carcinogen

antiseptic1488 [7]

100. ppb of chcl3 in drinking water means 100 g of CHCl3 in 1,000,0000,000 g of water

Molarity, M

M = number of moles of solute / volume of solution in liters

number of moles of solute = mass of CHCl3 / molar mass of CHCl3

molar mass of CHCl3 = 119.37 g/mol

number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol

using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters

M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer

Molality, m

m = number of moles of solute / kg of solvent

number of moles of solute = 0.838

kg of solvent = kg of water = 1,000,000 kg

m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer

mole fraction of solute, X solute

X solute = number of moles of solute / number of moles of solution

number of moles of solute = 0.838

number of moles of solution = number of moles of solute + number of moles of solvent

number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles

number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles

X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer

mass percent, %

% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =

% = 10 ^ - 6 % <------- answer

7 0

2 years ago

Read 2 more answers

Type the correct answer in the box. Express your answer to three significant figures.

satela [25.4K]

Answer:

The partial pressure of argon in the jar is 0.944 kilopascal.

Explanation:

Step 1: Data given

Volume of the jar of air = 25.0 L

Number of moles argon = 0.0104 moles

Temperature = 273 K

Step 2: Calculate the pressure of argon with the ideal gas law

p*V = nRT

p = (nRT)/V

⇒ with n = the number of moles of argon = 0.0104 moles

⇒ with R = the gas constant = 0.0821 L*atm/mol*K

⇒ with T = the temperature = 273 K

⇒ with V = the volume of the jar = 25.0 L

p = (0.0104 * 0.0821 * 273)/25.0

p = 0.00932 atm

1 atm =101.3 kPa

0.00932 atm = 101.3 * 0.00932 = 0.944 kPa

The partial pressure of argon in the jar is 0.944 kilopascal.

5 0

2 years ago