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aleksandr82 [10.1K]
3 years ago
9

Consider two gases, A and B, are in a container at room temperature.

Chemistry
2 answers:
prisoha [69]3 years ago
6 0
The answer is B. increase 
because when the pressure is doubled means the molecules are more close to each other so the reaction will be increase :)))
i hope this is helpful
have a nice day 
suter [353]3 years ago
4 0
The rate of reaction would increase because as pressure increases the molecules are more likely to bump into each other leading to a more likely hood of the molecules colliding properly to react leading to an increase in the reaction rate of the substance.
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A nuclear waste site. cesium-137 is a particularly dangerous by-product of nuclear reactors. it has a half-life of 30 years. it
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The mass decay rate is of the form
m(t) = m_{0} e^{-kt}
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k = the decay constant
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3 years ago
Sodium thiosulfate, na2s2o3, is used as a "fixer" in black and white photography. identify the reducing agent in the reaction of
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4 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
4 years ago
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