What is the expected charge on an ion of nitrogen? A. -3 B. +3 C. +8 D. +5

Based on the following chemical equation how many hydrogen atoms are present in the products side?

e-lub [12.9K]

Answer:

the answer is 6

Explanation:

there is 3 hydrogen molecules in NH3 and there's 2 molecules of NH3 so in total, there are 6 hydrogen molecules on the products side.

8 0

2 years ago

19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?

Alecsey [184]

Answer:

$m_{H_2O}=3.384gH_2O$

Explanation:

Hello,

In this case, the chemical reaction is:

$Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O$

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

$n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O$

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

$m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O$

Regards.

4 0

3 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago

The shape of the BF3 molecule is best described as

Mars2501 [29]

Answer:

The shape of the BF3 molecule is best described as trigonal planar.

Explanation:

The Lewis Structure for BF3 is like this:

_ _

| F | | F |

\ /

B

|

| F |

---

It forms three angles of 120° each. The bonds are in the same planar that's why it is trigonal planar and they are exactly the same.

Boron and Fluorine have 3 covalent bonds, produced by electronic promotion that enables the 2py and 2pz orbitals, leaving an electron to pair in the 2px. So boron will have 3 possible electrons to pair in 2s1, 2px and 2py, remember that electronic configuration for B is 1s2, 2s2, 2p1

By hybridization between the orbitals 2s2 and 2p1, the electrons of F, can joined to make the covalent bond. The new B configuration is 1s2, 2s1, 2px1, 2py1 (these last three, hybrid orbitals)

5 0

3 years ago

AV 5. How many grams are in 0.52 moles of H3PO4?

Hoochie [10]

Answer:

50.96g

Explanation:

Given parameters:

Number of moles of H₃PO₄ = 0.52moles

Unknown:

Mass of the compound = ?

Solution:

To find the mass of the compound:

Mass = number of moles x molar mass of H₃PO₄

Molar mass of H₃PO₄ = 3(1) + 31 + 4(16) = 98g/mol

Mass = 0.52 x 98 = 50.96g

3 0

3 years ago