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2 years ago

What is the mass of a sample of NH3 containing 7.20 × 1024 molecules of NH3? 161 grams 187 grams 203 grams 214 grams

Chemistry

2 answers:

Shkiper50 [21]2 years ago

7 0

Answer:

203 grams

Explanation:

It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).

Using cross multiplication:

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 7.2 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

∴ The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.

inn [45]2 years ago

3 0

Answer:

this is correct^

Explanation:

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Phosphoric acid has a pka of 2.1. at what ph will 75% of phosphoric acid be in the conjugate base form?

Colt1911 [192]

Answer:

pH is 2.58

Explanation:

The pH of the buffer made from the mixture of phosphoric acid (H₃PO₄) and its conjugate base form (H₂PO₄⁻) follows the Henderson-Hasselbalch equation:

pH = pka + log [H₂PO₄⁻] /[H₃PO₄]

If 75% of the buffer is in the conjugate base form (H₂PO₄⁻), 25% will be as H₃PO₄. Replacing to find pH:

pH = 2.1 + log [75%] /[25%]

pH = 2.58

pH is 2.58

8 0

2 years ago

Read 2 more answers

The elements X and Y combine in different ratios to form four different types of compounds: XY, XY2, XY3, and XY4. Consider that

Law Incorporation [45]

Answer:

Ratios in order of increasing value ; The ratio of the mass ratio of Y to X in XY2 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY3 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY4 to the mass ratio of Y to X in XY

1) Mass ratio = 3

2) Mass ratio = 2

3) Mass ratio = 4

Explanation:

The detailed and step by step calculation is shown in the attachment.

3 0

2 years ago

Calculate the nuclear binding energy for 5525mn in megaelectronvolts per nucleon (mev/nucleon). express your answer numerically

VARVARA [1.3K]

Answer:

8.533 Mev/nucleon

Explanation:

The given elements is:- $^{55}_{25}Mn$

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, the number of protons = 25

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

55 = 25 + Number of neutrons

Number of neutrons = 30

Mass of neutron = 1.008665 amu

Mass of proton = 1.007825 amu

Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron

Thus,

Calculated mass = (25*1.007277 + 30*1.008665) amu = 55.441875 amu

Actual mass = 54.9380451 amu

Mass defect = Δm = |54.9380451 - 55.441875| amu = 0.5038299 amu

The conversion of amu to MeV is shown below as:-

1 amu = 931.5 MeV

So, Energy = 0.5038299*931.5 MeV= 469.317552 MeV

Total number of nucleons in the atoms = 55

So, Energy = 469.317552/55 MeV/nucleon = 8.533 Mev/nucleon

5 0

2 years ago

1.00 g of a compound is combusted in oxygen and found to give 3.14g of CO2 and 1.29 g of H2O. From these data we can tell thatA.

lora16 [44]

Answer:

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula

Explanation:

Mass of CO2 obtained = 3.14 g

Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol

The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g

Mass of H2O obtained = 1.29 g

Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol

The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g

% by mass of carbon = 0.857/1 ×100 = 85.7 %

% by mass of hydrogen = 0.0717/1 × 100 = 7.17%

Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %

Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.

5 0

2 years ago

certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 40.00 gram sample of the alcohol pro

seraphim [82]

Answer:

The answer to your question is: C₂H₆O₁ = C₂H₆O

Explanation:

Data

CxHyOz

mass = 40 g produced 76.40 g of CO2

46.96 g of H2O

Empirical formula = ?

Process

CxHyOz + O2 ⇒ CO2 + H2O

44g of CO2 -------------------- 12 g of Carbon

76.40 g of CO2 --------------- x

x = 20.84 g of Carbon

12 g of Carbon --------------- 1 mol

20.84 g of C --------------- x

x = (20.84 x 1) / 12

x = 1.74 mol of Carbon

18 g of H2O -------------------- 2 g of H

46.96 g of H2O -------------- x

x = (46.96 x 2) / 18

x = 5.22 g of H

1 g of H ------------------------ 1 mol of H

5.22 g of H ------------------- x

x = 5.22 mol of H

Mass of Oxygen = 40 - 20.84 - 5.22g

= 13.94 g

16 g of Oxygen ---------------- 1 mol

13.94 g of O -------------------- x

x = 0.87 mol of O

Divide by the lowest number of moles

Carbon = 1.74 / 0.87 = 2

Hydrogen = 5.22 / 0.87 = 6

Oxygen = 0.87 / 0.87 = 1

Empirical formula

C₂H₆O₁ = C₂H₆O

3 0

2 years ago