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3 years ago

A sample of gas occupies 555mL at 175 cm Hg and a temperature of 37.0C. What is the new volume (in mL) at STP?

Chemistry

1 answer:

miv72 [106K]3 years ago

5 0

<u>Answer:</u> The new volume at STP is 1125.43 mL.

<u>Explanation:</u>

STP conditions:

Pressure of the gas is 760 mmHg and temperature of the gas is 273 K

To calculate the volume when temperature and pressure are changed, we use the equation given by combined gas law. The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

Conversion factor: 1 cm = 10 mm

$P_1=175cmHg=1750mmHg\\V_1=555mL\\T_1=37^oC=[37+273]=310K\\P_2=760mmHg\\V_2=?mL\\T_2=273K$

Putting values in above equation, we get:

$\frac{1750mmHg\times 555mL}{310K}=\frac{760mmHg\times V_2}{273K}\\\\V_2=1125.43mL$

Hence, the new volume at STP is 1125.43 mL.

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What is evidence of a physical change. A.ice melts B.silver tarnishes C.biycicle rusts

timama [110]

C. Physical Change.

3 0

3 years ago

What would the pOH of a solution be if the pH is 9.4

Alika [10]

<h2>Answer:</h2>

pOH = 4.6

<h2>Explanations:</h2>

The sum of pH and pOH of a solution is 14as shown:

$pH+pOH=14$

Given the following parameter

pH = 9.4

Substitute the given parameters into the formula to have:

$\begin{gathered} 9.4+pOH=14 \\ pOH=14-9.4 \\ pOH=4.6 \end{gathered}$

Hence the pOH of the solution of pH of 9.4 is 4.6

4 0

1 year ago

If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the part

lawyer [7]

This is an incomplete question, here is a complete question.

Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

$HN_3(g)\rightarrow N_2(g)+H_2(g)$

If 3.0 atm of pure HN₃ (g) is decomposed initially,what is the final total pressure in the reaction container? Whatare the partial pressures of nitrogen and hydrogen gas? Assume thatthe volume and temperature of the reaction container are constant.

Answer : The partial pressure of $N_2$ and $H_2$ gases are, 4.5 atm and 1.5 atm respectively.

Explanation :

The given unbalanced chemical reaction is:

$HN_3(g)\rightarrow N_2(g)+H_2(g)$

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the $HN_3$ and the coefficient '3' put before the $N_2$ then we get the balanced chemical equation.

The balanced chemical reaction will be,

$2HN_3(g)\rightarrow 3N_2(g)+H_2(g)$

As we are given:

The pressure of pure $HN_3$ = 3.0 atm

$p_{Total}=2\times p_{HN_3}=2\times 3.0atm=6.0atm$

From the reaction we conclude that:

Number of moles of $N_2$ = 3 mol

Number of moles of $H_2$ = 1 mol

Now we have to calculate the mole fraction of $N_2$ and $H_2$

$\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75$

and,

$\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25$

Now we have to calculate the partial pressure of $N_2$ and $H_2$

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 6.0 atm

$X_i$ = mole fraction of gas

$p_{N_2}=X_{N_2}\times p_T$

$p_{N_2}=0.75\times 6.0atm=4.5atm$

and,

$p_{H_2}=X_{H_2}\times p_T$

$p_{H_2}=0.25\times 6.0atm=1.5atm$

Thus, the partial pressure of $N_2$ and $H_2$ gases are, 4.5 atm and 1.5 atm respectively.

8 0

3 years ago

2. The last electron in arsenic occupies a 4p level. What block

Romashka-Z-Leto [24]

Answer:

Arsenic is present in p block and group fifteen of periodic table.

Explanation:

The atomic number of Arsenic is 33.

According to the Aufbau principal in ground state of elements electron first occupy the lower energy level then fill the higher energy levels. We know that there four subshells s, p, d and f. The maximum number of electrons in these subshells can be calculated by following formula:

2 (2l +1 )

and l = 0,1,2,3,....

maximum numbers of electrons in s subshell are,

l=0

2 ( 2(0) + 1)

so maximum electrons in s subshell are 2.

maximum numbers of electrons in p subshell are,

l = 1

2 ( 2(1) + 1)

2( 2 + 1)

so maximum electrons in p subshell are 6.

maximum numbers of electrons in d subshell are,

l = 2

2 ( 2(2) + 1)

2( 4 + 1)

so maximum electrons in d subshell are 10.

maximum numbers of electrons in f subshell are,

l = 3

2 ( 2(3) + 1)

2( 6 + 1)

so maximum electrons in f subshell are 14.

Electron first fill 1s subshell then 2s subshell and in this way they goes to higher energy levels.

Electronic configuration of Arsenic:

1s2 2s2 2p6 3s2 3p6 4s2 3d10, 4p3

The last electron is present in p subshell that's way Arsenic is present in p block of periodic table.

P block elements are non-metals, metals and metalloids. These are thirty five elements. The P-block elements are present on right side of periodic table. There valance electrons are present in P orbital. The p-block metals are shiny and good conductor of heat and electricity. These metal lose the electron which is accept by non metals and form ionic bond. They have high melting points.

Metalloids includes boron, silicon, germanium, arsenic, antimony and tellurium. Metalloids contain both the properties of metals and non metals, Some metalloids are toxic like arsenic.

Most of p-block elements are non metals. They are bad conductor of heat and electricity and have low boiling points.

6 0

3 years ago

The half-life of carbon-14 is 5,730 years. Dating organic material by looking for C-14 can't be accurately done after 50,000 yea

ki77a [65]

Answer:

0.010g of C-14 would be later after 50,000 years

Explanation:

The kinetics of radioactive decay follows the equation:

Ln (N / N₀) = -kt

<em>Where N could be taken as mass after time t, </em>

<em>N₀ initial mass = 4.30g;</em>

k is rate constant = ln 2 / t(1/2)

<em>= ln 2 / 5730years = 1.2097x10⁻⁴ years ⁻¹</em>

<em />

Replacing:

Ln (N / 4.30g) = -1.2097x10⁻⁴ years ⁻¹ * 50000 years

N / 4.30 = 2.36x10⁻³

N =

<h3>0.010g of C-14 would be later after 50,000 years</h3>

7 0

2 years ago