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lyudmila [28]
2 years ago
5

A.) What is the mass in g of 5 mols of Mg?

Chemistry
1 answer:
Dovator [93]2 years ago
6 0

Answer:

Mass of Mg  = 120 g

1.23 moles of Cu

Explanation:

A)

Given data:

Number of moles of Mg = 5 mol

Mass of Mg = ?

Solution:

Number of moles = mass/molar mass

Molar mass of magnesium is 24 g/mol.

5 mol = mass/ 24 g/mol

Mass = 5 mol × 24 g/mol

Mass = 120 g

B) Given data:

Number of atoms of Cu = 7.4 ×10²³ atom

Number of moles of Cu = ?

Solution:

Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

7.4 ×10²³ atom × 1 mol / 6.022 × 10²³ atom

1.23 moles of Cu

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OverLord2011 [107]

The answer is 0.405 M/s

- (1/3) d[O2]/dt = 1/2 d[N2]/dt

- d[O2]/dt = 3/2 d[N2]/dt

- d[O2]/dt = 3/2 × 0.27

- d[O2]/dt = 0.405 mol L^(-1) s^(-1)

5 0
2 years ago
In the laboratory you dissolve 19.4 g of potassium acetate in a volumetric flask and add water to a total volume of 125 mL. What
aleksandrvk [35]

The molarity of the potassium acetate solution given the data is 1.584 M

<h3>What is molarity? </h3>

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of CH₃COOK</h3>
  • Mass of CH₃COOK = 19.4 g
  • Molar mass of CH₃COOK = 98 g/mol
  • Mole of CH₃COOK =?

Mole = mass / molar mass

Mole of CH₃COOK = 19.4 / 98

Mole of CH₃COOK = 0.198 mole

<h3>How to determine the molarity of CH₃COOK</h3>
  • Mole of CH₃COOK = 0.198 mole
  • Volume = 125 mL = 125 / 1000 = 0.125 L
  • Molarity of CH₃COOK = ?

Molarity = mole / Volume

Molarity of CH₃COOK = 0.198 / 0.125

Molarity of CH₃COOK = 1.584 M

Learn more about molarity:

brainly.com/question/15370276

#SPJ4

5 0
2 years ago
What is the balanced equation for H3PO4=H4P2O7+H2O
podryga [215]

Answer:

2H3PO4=H4P2O7+H2O

Explanation:

2H3PO4=H4P2O7+H2O

5 0
3 years ago
A sample of gas occupies 7.80 liters at 425°C? What will be the volume of the gas at 35°C if the pressure does not change?
balandron [24]
From ideal gas equation PV = nRT, V/T = nR/P ==> V/T = constant. Therefore V1/T1 = V2/T2 ==> 7.8/698 = V2/308. V = 3.44L {TEMPERATURE IN KELVIN = 273 + 425 AND 35 = 698 AND 308}
4 0
2 years ago
A field worker is exposed to a xylene for a duration of 8 weeks at 40 hrs/wk. The concentration of xylene in the workplace is 40
Andrej [43]

Answer:

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

Explanation:

Number of hours worker exposed to xylene = 40 hr/week\times 8 week = 320 hours

The concentration of xylene in the workplace =40 \mu g/m^3

The worker is inhaling air at a rate of 0.9 m^3/hr.

Amount xylene inhaled by worker in an hour :

= 40\mu g/m^3\times 0.9 m^3/hr=36 \mu g/hr

Amount xylene inhaled by worker in 320 hours:

36 \mu g/hr\times 320 hr=11,520 \mu g=11,520\times 0.001 mg=11.520 mg

1 μg = 0.001 mg

Amount xylene inhaled by worker in 320 hours = 11.520 mg

1 day = 24 hours

Amount xylene inhaled by worker in 1 day:

\frac{24}{320}\times 11.520 mg=0.864 mg

Assuming 70 kg body mass, the chronic daily intake of xylene :

\frac{0.864 mg/day}{70 kg}=0.01234 mg/ kg day\approx 0.012 mg/ kg day

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

5 0
3 years ago
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