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Leni [432]
2 years ago
7

What volume would 0.435 moles of hydrogen gas, h2, occupy at stp?

Chemistry
2 answers:
faust18 [17]2 years ago
3 0
The conversion factor for volume at STP is \frac {1mol}{22.4L} or \frac {22.4L}{1mol}. Since we want volume, we would use \frac {22.4L}{1mol}. We conclude with the following calculations:

0.435mol H_{2} * \frac {22.4LH_{2}}{1molH_{2}} = 9.744L

The answer is 9.744L H2
zmey [24]2 years ago
3 0

Answer: The volume occupied by 0.435 moles of H_2 will be 9.7 Liters.

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

Thus if 1 mole of H_2  occupies 22.4 L at STP

0.435 moles of H_2  occupies=\frac{22.4}{1}\times 0.435=9.8L L at STP

The volume occupied by 0.435 moles of H_2 will be 9.7 Liters.

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How many molecules of glucose are in 1 l of a 100 mm glucose solution?
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Answer is: 6.022·10²² molecules of glucose.

c(glucose) = 100 mM.

c(glucose) = 100 · 10⁻³ mol/L.

c(glucose) = 0.1 mol/L; concentration of glucose solution.

V(glucose) = 1 L; volume of glucose solution.

n(glucose) = c(glucose) · V(glucose).

n(glucose) = 0.1 mol/L · 1 L.

n(glucose) = 0.1 mol; amount of substance.

N(glucose) = n(glucose) · Na (Avogadro constant).

N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.

N(glucose) = 6.022·10²².

6 0
2 years ago
In a chemical reaction that takes place at a fixed pressure and volume, the enthalpy change (ΔH) is –585 kJ/mol. Will this react
Molodets [167]

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5 0
2 years ago
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What would have happened to your results if during the dehydration some of the copper (ii) sulfate splatter out of the crucible-
lidiya [134]

Answer : The results would show more amount of water in the hydrated sample.

Explanation :

The amount of water of crystallization can be found by taking the masses of hydrated copper sulfate and anhydrous copper sulfate.

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4 0
2 years ago
In a group assignment, students are required to fill 10 beakers with 0.720 M CaCl2. If the molar mass of CaCl2 is 110.98 g/mol a
8_murik_8 [283]
The answer is 200 g.

If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M 
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So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be  79.90 g.

Now, we need to prepare ten beakers with 250 mL of solutions:
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79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
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8 0
2 years ago
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Figure 1. (a) Regular object of metal blocks with the same width, length, and height. (B) An irregular
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