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slega [8]
3 years ago
15

50 POINTS! PLEASE HELP!A gas in a balloon at constant pressure has a volume of 120.0mL at -12.30C. What is its volume at 197.00C

?
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
4 0

Answer:

Final volume=V₂ = 216.3 mL

Explanation:

Given data:

Initial volume = 120.0 mL

Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)

Final volume = ?

Final temperature = 197.0 °C (197+273 = 470 K)

Solution:

We will apply Charles Law to solve the problem.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 120 mL × 470 K /260.7K

V₂ = 56400 mL.K /260.7K

V₂ = 216.3 mL

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professor190 [17]

Answer:  

6.2 g  

Explanation:  

In a first-order decay, the formula for the amount remaining after <em>n</em> half-lives is  

N = \frac{N_{0}}{2^{n}}  

where  

<em>N</em>₀ and <em>N</em> are the initial and final amounts of the substance  

1. Calculate the <em>number of half-lives</em>.  

If t_{\frac{1}{2}} = \text{3.8 da}  

n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}

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4 0
3 years ago
WILL GIVE BRAINLIEST
Sav [38]

Answer:

the answer is A

Explanation:

5 0
4 years ago
Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i
g100num [7]
It's simple, just follow my steps.

1º - in 1 L we have 0.0100~g of BaCO_3

2º - let's find the number of moles.

\eta=\frac{m}{MM}

\eta=\frac{0.0100}{197.3}

\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}

3º - The concentration will be

C=5.07\times10^{-5}~mol/L

But we have this reaction

BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}

This concentration will be the concentration of Ba^{2+}~~and~~CO_3^{2-}

K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}

considering [BaCO_3]=1~mol/L

K_{sp}=[Ba^{2+}][CO_3^{2-}]

and

[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L

We can replace it

K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})

K_{sp}\approx25.70\times10^{-10}

Therefore the K_{sp} is:

\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}
4 0
4 years ago
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