Calculate: (7.0 x 10^5) + (2.5 x 10^4) =

Ammonia and oxygen react to produce nitric oxide and water. Which of the following chemical equations describes this reaction an

aksik [14]

Answer:

Option B. 4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Explanation:

The reaction between ammonia, NH3 and oxygen, O2, will produce nitric oxide and water as shown below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

Now, let us balance the equation.

This is illustrated below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

There are 3 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of NH3 and 6 in front of H2O as shown below:

4NH3(g) + O2(g) —> NO(g) + 6H2O(g)

There are 4 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 4 in front of NO as shown below:

4NH3(g) + O2(g) —> 4NO(g) + 6H2O(g)

Now, the are 2 atoms of O on the left side and a total of 10 atoms on the right side. It can be balance by putting 5 in front of O2 as shown below:

4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Now, we can see that the equation is balanced.

Therefore, option B gives the balanced chemical equation for the reaction.

5 0

3 years ago

What is the pH of a 0.300 M NH₃ solution that has Kb = 1.8 × 10⁻⁵ ? The equation for the dissociation of NH₃ is: NH₃ (aq) + H₂O

nalin [4]

Answer:

11.4

Explanation:

Step 1: Given data

Concentration of the base (Cb): 0.300 M

Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

$[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M$

Step 4: Calculate the pOH

We will use the following expression.

$pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6$

Step 5: Calculate the pH

We will use the following expression.

$pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4$

8 0

3 years ago

Which of the following is an oxidation half reaction

Ksju [112]

Answer:

Sn²⁺ → Sn⁴⁺ + 2e⁻

Explanation:

3 0

2 years ago

Read 2 more answers

Whats the formula for soduim hydrogen carbonate

Grace [21]

Answer:

NaHCO3

Explanation:

6 0

3 years ago

Read 2 more answers

How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o

Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

$Density (\rho)=\frac{Mass(m)}{Volume(V)}$

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,

$334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)$

$334x+x\times 11.976=29553.16$

345.976x = 29553.16

x = 85.4197 kg

Thus,

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

7 0

3 years ago