Nuclear fusion reactions occur in

Exit Calcium has an atomic number of 20. How many electrons of a calcium atom are involved in chemical bonding? A. 2 B. 8 C. 16

Colt1911 [192]

A. 2 is the answer let me know if im wrong sorry if i am.

6 0

3 years ago

Read 2 more answers

What is the percent by mass of water in MgSO4 • 7H2O? A. 51.1% B. 195% C. 56.0% D. 21.0%

lbvjy [14]

Answer : The correct option is, (A) 51.1%

Explanation :

Mass percent : It is defined as the mass of the given component present in the total mass of the compound.

Formula used :

$\text{Mass} \%H_2O=\frac{\text{Mass of }H_2O}{\text{Mass of }MgSO_4.7H_2O}\times 100$

First we have to calculate the mass of $H_2O$ and $MgSO_4.7H_2O$ .

Mass of $H_2O$ = 18 g/mole

Mass of $7H_2O$ = 7 × 18 g/mole = 126 g/mole

Mass of $MgSO_4.7H_2O$ = 246.47 g/mole

Now put all the given values in the above formula, we get the mass percent of $H_2O$ in $MgSO_4.7H_2O$ .

$\text{Mass} \%H_2O=\frac{126g/mole}{246.47g/mole}\times 100=51.1$

Therefore, the mass percent of $H_2O$ in $MgSO_4.7H_2O$ is, 51.1%

6 0

3 years ago

Calcium hydride reacts with water to form calcium hydroxide (aqueous) and hydrogen gas. Write a balanced chemical equation for t

DanielleElmas [232]

Answer:

$CaH_2(s) +2H_2O(l) \rightarrow Ca(OH)_2 (aq) + 2H_2 (g)$

Explanation:

$CaH_2(s) +H_2O(l) \rightarrow Ca(OH)_2 (aq) + H_2 (g)$

In this reaction, calcium is balanced in either side. No. of atoms of 'H' and 'O' are not balanced. To balance O atoms, multiply $H_2O$ by 2. Now the reaction becomes:

$CaH_2(s) +2H_2O(l) \rightarrow Ca(OH)_2 (aq) + H_2 (g)$

Now, to balance H atoms, multiply H2 by 2.

$CaH_2(s) +2H_2O(l) \rightarrow Ca(OH)_2 (aq) + 2H_2 (g)$

Now, each element is balanced.

8 0

3 years ago

At 300K, the pressure inside a gas-filled container is 500 kPa. If the pressure inside decreases to 100 kPa, but the volume rema

alukav5142 [94]

PV = nRT

⇒P1/T1 = P2/T2
⇒500/300 = 100/T2
⇒ T2 = 100×300/500 = 60K

Answer is D

3 0

4 years ago

Read 2 more answers

Nicotine, a component of tobacco, is composed of c, h, and n. a 4.200-mg sample of nicotine was combusted, producing 11.394 mg o

Rudiy27

Answer:

Empirical Formula = C₅H₇N₁

Solution:

Data Given:

Mass of Nicotine = 4.20 mg = 0.0042 g

Mass of CO₂ = 11.394 mg = 0.011394 g

Mass of H₂O = 3.266 mg = 0.003266 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.011394 ÷ 0.0042) × (12 ÷ 44) × 100

%C = (2.7128) × (12 ÷ 44) × 100

%C = 2.7128 × 0.2727 × 100

%C = 73.979 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.003266 ÷ 0.0042) × (2.02 ÷ 18.02) × 100

%H = (0.7776) × (2.02 ÷ 18.02) × 100

%H = 0.7776 × 0.1120 × 100

%H = 8.709 %

%N = 100% - (%C + %H)

%N = 100% - (73.979 % + 8.709%)

%N = 100% - 82.688%

%N = 17.312 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 73.979 ÷ 12.01

Moles of C = 6.1597 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 8.709 ÷ 1.01

Moles of H = 8.6227 mol

Moles of N = %N ÷ At.Mass of O

Moles of N = 17.312 ÷ 14.01

Moles of N = 1.2356 mol

Step 3: Find out mole ratio and simplify it;

C H N

6.1597 8.6227 1.2356

6.1597/1.2356 8.6227/1.2356 1.2356/1.2356

4.985 6.978 1

≈ 5 ≈ 7 1

Result:

Empirical Formula = C₅H₇N₁

6 0

3 years ago