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marin [14]
3 years ago
8

1. A gas has a volume of 45 L at a pressure of 140 kPa. What is the volume when

Chemistry
1 answer:
pishuonlain [190]3 years ago
7 0

Answer:

The new volume of the gas is 21 L.

Explanation:

Volume of a gas is inversely proportional to its pressure at constant temperature such that,

V\propto \dfrac{1}{P}

or

P_1V_1=P_2V_2

We have,

V_1=45\ L\\\\P_1=140\ kPa \\\\P_2=300\ kPa

It is required to find V₂. Using above law or Boyle's law such that :

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{140\times 45}{300}\\\\V_2=21\ L

So, the new volume of the gas is 21 L.

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The answer is syncline
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3 years ago
4Cr(s)+3O2(g)→2Cr2O3(s) calculate how many grams of the product form when 21.4 g of O2 completely reacts
weqwewe [10]

Answer:

= 67.79 g

Explanation:

The equation for the reaction is;

4Cr(s)+3O2(g)→2Cr2O3(s)

The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;

moles O2 = 21.4 g / 32 g/mol

                =0.669 moles

Using mole ratio, we get the moles of Cr2O3;

moles Cr2O3 = 0.669 x 2/3

                       =0.446 moles

but molar mass of Cr2O3 is 151.99 g/mol

Hence,

The mass Cr2O3 = 0.446 mol x 151.99 g/mol

                            <u> = 67.79 g </u>

6 0
3 years ago
Be<br><br> Name the element.<br><br> Number of shells?<br><br> Valence electrons?
Vedmedyk [2.9K]

Answer:

Name the element: Beryllium

Number of shells:  4

Valence electrons: 2

Explanation:

4 0
3 years ago
Which of the following is used as a refrigerant?<br> methane<br> esters<br> freon<br> fluorine
zepelin [54]

Answer: I believe the answer would be methane

Explanation:

Hope this helps

6 0
3 years ago
If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did
user100 [1]

<u>Answer:</u>

<em>4.5 L water we have in litres (L).</em>

<em><u></u></em>

<u>Explanation:</u>

Q=m\times c \times \Delta T

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

So,

Volume of water = mass/density

\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$

=4.5 L (Answer)

6 0
4 years ago
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