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ivann1987 [24]
3 years ago
12

A student is monitoring the pressure of a gas in a rigid container. The amount of gas particles and the volume of the container

are held constant. How will the pressure be affected if the container is cooled over time?
Chemistry
2 answers:
daser333 [38]3 years ago
7 0

Answer:

the gas pressure will decrease

Explanation:

saveliy_v [14]3 years ago
3 0
Chances is it will drop because in the cold gas cannot expand or rise while in the heat it expands and rises 
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What are the coefficients to balance the following equation?<br><br> ba+br2=babr2
kaheart [24]

Answer:

Its already balanced

Explanation:

Ba (barium) starts with and ends with 1. Br (bromine) starts with 2 and ends with 2.

6 0
3 years ago
Read 2 more answers
The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be
Stels [109]

Explanation:

1) Initial mass of the Cesium-137=N_o= 180 mg

Mass of Cesium after time t = N

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

Half life of the cesium-137 = t_{1/2}=30 years[p/tex]where,&#10;[tex]N_o = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

t_{\frac{1}{2}} = half life of the isotope

\lambda = rate constant

N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}

Now put all the given values in this formula, we get

N=180mg\times e^{-\frac{0.693}{30 years}\times t}

Mass that remains after t years.

N=180 mg\times e^{0.0231 year^{-1}\times t}

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

N_o=180 mg

t_{1/2}= 30 yeras

N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

N_o=180 mg

t_{1/2}= 30 yeras

N = 1 mg

t = ?

1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}

\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0
3 years ago
NEED HELP FAST!!
koban [17]

Answer:

I'm pretty sure that's right.

Explanation:

Download pdf
8 0
3 years ago
Determine the [h3o+] of a 0.210 m solution of formic acid.
Nataly [62]
When the Pka for formic acid = 3.77
and Pka = -㏒ Ka 
   3.77 = -㏒ Ka
∴Ka = 1.7x10^-4 

when Ka = [H+][HCOO-}/[HCOOH]

when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2      by solving this equation for X

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
       = -㏒ 0.0059
       = 2.23 

3 0
3 years ago
Equation balancing
meriva

<u>Answer:</u>

<u>For a:</u> The balanced equation is 2S(s)+3O_2(g)\rightarrow 2SO_3(g)

<u>For c:</u> The balanced equation is 2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

<u>Explanation:</u>

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

  • <u>For (a):</u>

The given unbalanced equation follows:

S(s)+O_2(g)\rightarrow SO_3(g)

To balance the equation, we must balance the atoms by adding 2 infront of both S(s) and SO_3 and 3 in front of O_2

For the balanced chemical equation:

2S(s)+3O_2(g)\rightarrow 2SO_3(g)

  • <u>For (b):</u>

The given balanced equation follows:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

The given equation is already balanced.

  • <u>For (c):</u>

The given unbalanced equation follows:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)

To balance the equation, we must balance the atoms by adding 2 infront of H_2O(l)

For the balanced chemical equation:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

  • <u>For (d):</u>

The given balanced equation follows:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

The given equation is already balanced.

4 0
3 years ago
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