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3 years ago

The volume of a sample of gas is 0.5 L and the pressure is 0.98 atm. The volume is increased to 1.0 L. Show the set

Chemistry

1 answer:

meriva3 years ago

5 0

Given,

P1 = 0.98 atm

V1 = 0.5 L

V2 = 1.0 L

P2 = ?

Solution,

According to Boyle's Law,

P1V1 = P2V2

0.98 × 0.5 = 1.0 × P2

P2 = 0.98 × 0.5 × 1.0

P2 = 0.49 atm

Answer - The new pressure is 0.49 atm.

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Erika is asked to create a model showing a negatively charged Carbon ion. Which model should she create?

OverLord2011 [107]

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2 years ago

The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.

user100 [1]

Answer:

c 43.38 g

Explanation:

The reaction between MnO2 and HCl can be represented by the following balanced equation:

MnO2 + HCl ---> Cl2 + MnCl2 + H2O

From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.

For the fact that HCl gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess, while the limiting reagent should be MnO2 .

Thus, the theoretical yield of Cl2 will be 60.25 g.

By definition, the percentage yield is given by

% Yield = (Actual Yield) / (Theoretical Yield),

This can be simplified to

Actual Yield = % Yield * Theoretical Yield

Plugging in the given values we have

Actual Yield = 72% * 60.25 = 43.38 g

5 0

3 years ago

One molecule of a compound weighs 2.03x10-22 g. what is the molar mass of that compound?

stealth61 [152]

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3 years ago

Read 2 more answers

Calculate the change in entropy if Br2(l) is converted to Br2(g).

Alik [6]

Ans: The entropy change for the given reaction is 93.3 J/K

Given reaction:

Br2(l) → Br2(g)

ΔS = ∑n(products)S⁰(products) - ∑n(reactants)S⁰(reactants)

= 1 mole* S°(Br2(g)) - 1 mole*S°(Br2(l))

= 1 mole *245.5 J/mol-K - 1 mole*152.2 J/mol-K

= 93.3 J/K

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3 years ago

12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?

Kaylis [27]

Answer:

$115.625^{\circ}\text{F}$

Explanation:

$m_1$ = First mass of water = 12 oz

$m_2$ = Second mass of water = 20 oz

$\Delta T_1$ = Temperature difference of the solution with respect to the first mass of water = $(T-75)^{\circ}\text{F}$

$\Delta T_2$ = Temperature difference of the solution with respect to the second mass of water = $(T-75)^{\circ}\text{F}$

c = Specific heat of water

As heat gain and loss in the system is equal we have

$m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}$

The final temperature of the solution is $115.625^{\circ}\text{F}$ .

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2 years ago