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3 years ago

Carbon dioxide is produced when dilute hydrochloric acid reacts with

Chemistry

1 answer:

user100 [1]3 years ago

8 0

Carbon dioxide is created when hydrolic is mixed with calcium carbonate.

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If i have 4 moles of a gas ata pressure of 5.2 atm and a volume of 15 liters, what is the temperature?

kompoz [17]

Answer:

237.5 K.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 5.2 atm).

V is the volume of the gas in L (V = 15.0 L).

n is the no. of moles of the gas in mol (n = 4.0 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = ??? K).

∴ T = PV/nR = (5.2 atm)(15.0 L)/(4.0 mol)(0.0821 L.atm/mol.K) = 237.5 K.

5 0

3 years ago

What pressure (in kilopascals) is exerted by 4.20 moles of Xenon gas in a 15.0 L container at 280.0 K?

aniked [119]

Answer: the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa

Explanation:

The question requires us to calculate the pressure, in kPa, connsidering the following information:

number of moles = n = 4.20mol

volume of gas = V = 15.0L

temperature of gas = T = 280.0 K

We can use the equation of ideal gases to calculate the pressure of the gas, as shown by the rearranged equation below:

$PV=nRT\rightarrow P=\frac{nRT}{V}$

Since the volume was given in L and the question requires us to calculate the pressure in kPa, we can use R in units of L.Pa/K.mol:

R = 8314.46 L.Pa/K.mol

Applying the values given by the question to the rearranged equation above, we'll have:

$\begin{gathered} P=\frac{nRT}{V} \\ \\ P=\frac{(4.20mol)\times(8314.46L.Pa/K.mol)\times(280.0K)}{(15.0L)}=652\times10^3Pa \end{gathered}$

Therefore, the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa.

8 0

1 year ago

Give an example of two elements that would form a covalent bond

Tamiku [17]

Hydrogen carbon and oxygen commonly form covalent bonds.

HOPE THIS HELPS!

4 0

3 years ago

What are the number of atoms in each Chemical Formula

podryga [215]

You can search that up online it’s not that hard but good luck !!

6 0

3 years ago

Lina20 [59]

Answer:

Please see attachment

Explanation:

Please see attachment

6 0

3 years ago