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3 years ago

What is the volume of 2.00 moles of ideal gas at 25'c and 121.59 kpa of pressure

Chemistry

1 answer:

katrin [286]3 years ago

6 0

Answer:

40.73 L.

Explanation:

We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 121.59 kPa/101.325 = 1.2 atm).

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

n is the no. of moles of the gas in mol (n = 2.0 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K (T = 25°C + 273 = 298 K).

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What is the major product in this reaction

Butoxors [25]

Answer:

I think option A is right answer

4 0

3 years ago

Which best compares a molecule that has a trigonal planar shape with one that has a trigonal pyramidal shape?

Juli2301 [7.4K]

Answer:

B. They both contain three atoms around the central atom.

Explanation:

Do the unit test on edg20 and got it right!

7 0

3 years ago

For the following reaction, 22.0 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas . nitrogen mono

mash [69]

NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete

<h3>What is a limiting reagent?</h3>

The reactant that is entirely used up in a reaction is called as limiting reagent.

The reaction:

$2NO(g) +2H_2(g)$ → $N_2 +2H_2O$

Moles of nitrogen monoxide

Molecular weight: $M_(_N_O_)$ =30g/mol

$n_(_N_O_) =\frac{mass}{molar \;mass}$

$n_(_N_O_) =\frac{22.0}{30g/mol}$

$n_(_N_O_) = 0.73 mol$

Moles of hydrogen

Molecular weight: $M_(_H_2_)$ =30g/mol

$n_(_H_2_) =\frac{mass}{molar \;mass}$

$n_(_H_2_) =\frac{5.80g}{2g/mol}$

$n_(_H_2_) = 2.9 mol$

Hydrogen gas is in excess.

NO is the limiting reagent.

The amount of the excess reagent remains after the reaction is complete.

$n_(_N_2_) =$ (2.9 mol- 0.73 mol NO x $\frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}$ ) x $\frac{2g \;of \;H_2}{mole \;of \;H_2}$

$n_(_N_2_) =$ 4.34 g

Learn more about limiting reagents here:

brainly.com/question/26905271

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5 0

1 year ago

How many hydrogen atoms are in 5.20 mol of ammonium sulfide?

Fofino [41]

1) Molecular formula of ammonium sulfide

(NH4)2 S

2) That means that there are 2*4 = 8 atoms of hydrogen in each molecule of ammoium sulfide, so in 5.20 mol of molecules will be 8 * 5.20 mol = 41.6 moles of atoms of hydrogen

3) To pass to number of atoms multiply by Avogadro's number: 6.022 * 10^23

41.6 moles * 6.022 * 10^23 atoms / mol = 250.5 * 10^23 = 2.50 * 10^25 atoms

Answer: 2.50 * 10^25

8 0

3 years ago

Calculate the molarity of a salt solution with a volume of 0.250L that contains 0.70 mol of NaCl. (SHOW WORK)

Kazeer [188]

Answer:

= 0.28M

Explanation:

data:

volume = 0.250 L

= 0.250dm^3 ( 1litre = 1dm^3)

moles = 0.70 moles

Solution:

molarity = $\frac{no. of moles}{volume in dm^3}$

= 0.70 / 0.250

molarity = 0.28 M

5 0

2 years ago