7. Iron combines with 4.00 g of Copper (11) nitrate to form 6.01 g of Iron (I) nitrate and 0.400 g copper

Pls help

Simora [160]

Heat required to raise the temperature = 159.505 J

<h3>Further explanation</h3>

Given

c = specific heat of Beryllium = 1.825 J/g C

m = mass = 2.3 g

Δt = Temperature difference : 60 - 22 = 38 °C

Required

Heat required

Solution

Heat can be formulated

Q = m.c.Δt

Input the value :

Q = 2.3 x 1.825 x 38

Q = 159.505 J

8 0

2 years ago

Consider 2 Al + 6 HCl → 2 AlCl3 + 3 H2 , the reaction of Al with HCl to produce hydrogen gas. What is the pressure of H2 if the

Margaret [11]

Answer : The pressure of hydrogen gas is 8.96 atm.

Explanation :

The given balanced chemical reaction is:

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

From the balanced chemical reaction we conclude that,

As, 2 moles of Al react to give 3 moles of $H_2$ gas

So, 4.50 moles of Al react to give $\frac{3}{2}\times 4.50=6.75$ moles of $H_2$ gas

Now we have to calculate the moles of $H_2$ gas when percent yield is 75.4.

$\text{The moles of }H_2=75.4\% \times 6.75=\frac{75.4}{100}\times 6.75=5.09moles$

Now we have to calculate the pressure of $H_2$ gas.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of hydrogen gas = ?

V = Volume of the hydrogen gas = 14.0 L

n = number of moles of hydrogen gas = 5.09 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of hydrogen gas = 300 K

Putting values in above equation, we get:

$P\times 14.0L=5.09mol\times 0.0821L.atm/mol.K\times 300K\\\\P=8.96atm$

Therefore, the pressure of hydrogen gas is 8.96 atm.

3 0

3 years ago

What is the ΔG for the following reaction at 25°C?

ikadub [295]

$\Delta G_{global}=\Delta G_{products}-\Delta G_{reagents}\\\\
\text{We should consider the stoichiometry:}\\\\
\Delta G=2\Delta G_{NO_2}-\Delta G_{N_2O_4}\\\\
\Delta G=2\cdot51.8-98.28=103.6-98.28\\\\
\boxed{\Delta G=5.32~kJ}$

4 0

3 years ago

Read 2 more answers

A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0

3 years ago

Match each event with its discription

Andre45 [30]

Answer:

full moon to A

new moon to B

solar eclipse to D

lunar eclipse to C

8 0

3 years ago