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3 years ago

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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?

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 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = 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:

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