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liberstina [14]
3 years ago
11

How many moles of gas does it take to occupy 25.6 liters at a pressure of 6.2 atmospheres and a temperature of 309.3 K?

Chemistry
1 answer:
Vilka [71]3 years ago
7 0

Answer:

45 moles.

Explanation:

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Cellular respiration

is the process that releases energy by breaking down glucose and other food molecules in the presence of oxygen

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A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo
lara [203]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M

6 0
3 years ago
Please answer a b c :)​
olganol [36]

Answer:

i think the letter b is the answer

Explanation:

sana makatulong

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2 years ago
For a laboratory investigation some students put a strip of shiny metal into a beaker of blue solution and then stored the beake
KatRina [158]

Answer:

b

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3 years ago
Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low
LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

4 0
3 years ago
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