All categories

3 years ago

Calculate the mass of silver bromide produced from 22.5 g of silver nitrate.

1 answer:

6 0

Equation is as follow,

 2 AgNO₃ + MgBr₂ → 2 AgBr + Mg(NO₃)₂

According to eq.

339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr
So,
22.5 g AgNO₃ will produce = X g of AgBr

Solving for X,
X = (22.5 g × 375.54 g) ÷ 339.74 g

X = 24.87 g of AgBr

You might be interested in

Please help me to answer this question. i have to submit tomorrow.

ratelena [41]

Answer:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

4 0

3 years ago

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 5.90 moles of magnesium perchlorate, Mg(ClO4)2.

Lady bird [3.3K]

Answer:

5.90, 11.8, 47.2

Explanation:

Let’s remove the parentheses and write the formula as MgCl₂O₈.

We see that 1 mol Mg(ClO₄)₂ contains 1 mol Mg atoms, 2 mol Cl atoms, and 8 mol O atoms.

∴ $\text{Moles of Mg atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{1 mol Mg atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{5.90 mol Mg atoms}$

$\text{Moles of Cl atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{2 mol Cl atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{11.8 mol Cl atoms}$

$\text{Moles of O atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{8 mol O atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{47.2 mol O atoms}$

∴ Mg, Cl, O = 5.90, 11.8, 47.2

4 0

3 years ago

The density of liquid mercury is 13.6 g/mL. What is its density in units of ? (2.54 cm = 1 in., 2.205 lb = 1 kg)

nalin [4]

Correct question

The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in3? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)

Answer:

$\rho0.4916\ lb/in^3$

Explanation:

Given that;-

The density = 13.6 g/mL

Also, 1 kg = 2.205 lb

1 kg = 1000 g

So, 1000 g = 2.205 lb

1 g = 0.002205 lb

Also,

1 in = 2.54 cm

1 in³ = 16.39 cm³

1 cm³ = 1 mL

So, 1 in³ = 16.39 mL

1 mL = 0.061 in³

The expression for the calculation of density is shown below as:-

$\rho=\frac{m}{V}$

Thus,

$\rho=\frac{13.6\ g}{1\ mL}=\frac{13.6\times 0.002205\ lb}{0.061\ in^3}=0.4916\ lb/in^3$

7 0

3 years ago

Lemon juice has a pH of 2. What is the hydrogen ion concentration of lemon juice?

saw5 [17]

PH = −log [H+] = − log [5.4 × 10−3] ≈ 2.27 or 2.3.
or basically 2

7 0

3 years ago

An element has two isotopes, A and B, with mass numbers of 32.0 and 33.0 respectively. The relative atomic mass is 32.3. The per

vlada-n [284]

Answer:

d) A - 70% B - 30%

Explanation:

If x is the abundance of A, and 1−x is the abundance of B, then:

x (32.0) + (1−x) (33.0) = 32.3

32x + 33 − 33x = 32.3

33 − x = 32.3

x = 0.7

The abundance of A is 70%, and the abundance of B is 30%.

7 0

3 years ago

Read 2 more answers