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frutty [35]
3 years ago
11

What human activity has caused the Increase of CO2 in the atmosphere?

Chemistry
1 answer:
Kaylis [27]3 years ago
4 0
Burning fossil fuels for energy.
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How long will it take for a 40.0 gram sample of I-131 (half-life = 8.040 days) to decay to 1/100 its original mass? ( PLEASE WIT
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A 0.539 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M(s)+H2SO4(aq)⟶MSO4(aq)+H2(g) A v
Charra [1.4K]

Answer:

The molar mass of the metal is 54.9 g/mol.

Explanation:

When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.

Patm = Pwater + PH₂

PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar

The pressure of H₂ is:

0.9762bar.\frac{1atm}{1.013bar} =0.9637atm

The absolute temperature is:

K = °C + 273 = 25°C + 273 = 298 K

We can calculate the moles of H₂ using the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{0.9637atm \times 0.249L }{(0.08206atm.L/mol.K)\times298K} =9.81 \times 10^{-3} mol

Let's consider the following balanced equation.

M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)

The molar ratio of M:H₂ is 1:1. So, 9.81  × 10⁻³ moles of M reacted. The molar mass of the metal is:

\frac{0.539g}{9.81 \times 10^{-3} mol} =54.9g/mol

4 0
3 years ago
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

  • 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

Xg = 9.3 g * 0.1 L / 1 L = 0.93 g

8 0
3 years ago
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