Please help me with this chem question!!

Sulfuric acid dissolves aluminum metal according to the following reaction:

Fiesta28 [93]

Answer:

$m_{H_2SO_4}=81.7gH_2SO_4$

$m_{H_2}=1.67gH_2$

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

$2Al (s) + 3H_2SO_4 (aq)\rightarrow Al _2(SO4)_3 (aq) + 3H_2 (g)$

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

$m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4$

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

$m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2$

Best regards.

3 0

3 years ago

An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit

Anon25 [30]

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of triprotic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL$

Putting values in above equation, we get:

$3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

$0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol$

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

7 0

2 years ago

When a 120 g sample of aluminum absorbs 9612 of heat energy, its temperature increases from 25°C to 115°C. Find the specific hea

lesantik [10]

<h3>Answer:</h3>

0.89 J/g°C

<h3>Explanation:</h3>

Concept tested: Quantity of heat

We are given;

Mass of the aluminium sample is 120 g
Quantity of heat absorbed by aluminium sample is 9612 g
Change in temperature, ΔT = 115°C - 25°C

= 90°C

We are required to calculate the specific heat capacity;

We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.

That is;

Q = m × c × ΔT

Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.

Specific heat capacity, c = Q ÷ mΔT

= 9612 J ÷ (120 g × 90°C)

= 0.89 J/g°C

Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C

4 0

3 years ago

5. What is the partial pressure of the remaining gas, if the total is 384 torr 2

alexgriva [62]

Answer: 83.11 torr

Explanation:

According to Dalton's Law of partial pressure, the total pressure of a mixture of gases is the sum of the pressure of each individual gas.

i.e Ptotal = P1 + P2 + P3 + .......

In this case,

Ptotal = 384 torr

P1 = 289 torr

P2 = 11.89 torr

P3 = ? (let the partial pressure of the remaining gas be Z)

Ptotal = P1 + P2 + Z

384 torr = 289 torr + 11.89 torr + Z

384 torr = 300.89 torr + Z

Z = 384 torr - 300.89 torr

Z = 83.11 torr

Thus, the partial pressure of the remaining gas is 83.11 torr.

5 0

3 years ago

I NEED HELP WITH #3 pleaseeee

Anit [1.1K]

Answer:

energy level

sub energy level

electrons in the orbital

Explanation:

3 0

3 years ago