How many grams of magnesium is needed to produce .0054g of hydrogen gas?

A sample of 2.45g aluminum oxide decomposes into 1.3g of aluminum and 1.15g of oxygen. What is the percentage composition of the

Vitek1552 [10]

Answer:

% $Al = 53.1%$ %

% $O = 46.9$ %

Explanation:

If we know the grams of a chemical compound in a specific reaction, it is possible to know the percentage of each atom that composes it.

For the Aluminum Oxide in this problem, we know its total weight and the grams of each component.

therefore we can determine the percentage ratio of its components through:

For Al

% $Al = \frac{mass of Al}{mass of Aluminium oxide} . 100$ %

% $Al = \frac{1,3 g}{2,45 g} . 100$ %

% $Al = 53.1%$ %

In the same way for oxygen

% $O = \frac{mass of O}{mass of Aluminium oxide} . 100$ %

% $O = \frac{1,5 g}{2,45 g} . 100$ %

% $O = 46.9$ %

5 0

3 years ago

Question 9 (1 point)

pantera1 [17]

Answer:

8 moles of C

Explanation:

From the question given above, the following equation was obtained:

3A + 2B —> 6C

From the equation above,

3 moles of A reacted to produce 6 moles of C.

Thus, the number of mole of C produced by reacting 4 moles of A can be obtained as follow:

From the equation above,

3 moles of A reacted to produce 6 moles of C.

Therefore, 4 moles of C will react to produce = (4 × 6)/3 = 8 moles of C

Thus, 8 moles of C can be obtained from the reaction of 4 moles of A with excess B

6 0

3 years ago

How many liters of oxygen are required to react completely with 1.2 liters of hydrogen to form water? 2H2(g) + O2(g) → 2H2(g)

Ira Lisetskai [31]

Answer:

$\boxed{\text{0.60 L}}$

Explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem.

Gases at the same temperature and pressure react in the same ratios as their coefficients in the balanced equation.

1. Write the chemical equation.

Ratio: 2 L 1 L

2H₂ + O₂ → 2H₂O

V/L: 1.2

2. Calculate the volume of O₂.

According to Gay-Lussac, 1 L of O₂ forms from 2 L of H₂.

Then, the conversion factor is (1 L O₂/2 L H₂).

$\text{Volume of O}_{2} = \text{1.20 L H}_{2}\times \dfrac{\text{1 L O}_{2}}{\text{2 L H}_{2}} = \textbf{0.60 L O}_{2}\\\\\text{You need }\boxed{\textbf{0.60 L of O}_{2}}$

4 0

3 years ago

3. A 650 mg sample is analyzed and found to contain

JulsSmile [24]

Answer:

7.85% of Hydrogen in the sample.

Explanation:

51/650 = .07846 x 100= 7.846% (round up)

6 0

4 years ago

Find the new concentration of a solution if 25.0 mL of water is added to 125.0 mL of 0.150 M NaCl solution. What is the final vo

lesantik [10]

Answer : The correct answer is New concentration = 0.125 M and final volume = 0.150 M

Given :

Molarity of NaCl solution = 0.150 M volume of NaCL solution = 125 mL

Volume of water added = 25 mL

1) Final volume = ? 2) New concentration = ?

1) To find the final volume :

New volume = volume of NaCl solution + volume of water added

= 125 mL + 25 mL

= 150 mL

Converting this final volume from mL to L

1 L = 1000 mL

$Final volume = 150 mL * \frac{1 L}{1000 mL }$

Final volume = 0.150 L

2) To find new concentration :

Following steps can be used :

a) To find mole of NaCl :

Mole of NaCl can be calculated using molarity formula . Molarity can be defined as mole of solute present in Liter of solution (volume ) .It represents Concentration of solution . It used unit as M or $\frac{mol}{L}$ . It can be expressed as :