All categories

3 years ago

Calculate the pressure in atmospheres of 0.5 mol of Helium ( He) at 20°C & occupying 2.50 L.

Chemistry

1 answer:

Morgarella [4.7K]3 years ago

7 0

Answer:

4.811 atm.

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 = ?? atm).

V is the volume of the gas in L (V = 2.5 L).

n is the no. of moles of the gas in mol (n = 0.5 mol).

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

T is the temperature of the gas in K (T = 20°C + 273 = 293 K).

You might be interested in

Examples of solid mixtures

elixir [45]

Salad, a bag of different colored pebbles or sand, etc

6 0

3 years ago

When an object falls toward the ground due to gravity, what type of energy becomes kinetic energy?

Goshia [24]

OPTION C

<h2>POTENTIAL ENERGY </h2>

THANK U

6 0

3 years ago

Read 2 more answers

Please help this is my chemistry work

Alja [10]

It’s deceleration hope that helps!

5 0

3 years ago

Read 2 more answers

Rick has taken a loan of $3,200 from the bank to buy new appliances. His first loan payment is due at the end of this month. He

marysya [2.9K]

Answer:

140

Explanation:

4 0

3 years ago

Read 2 more answers

When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

3 years ago