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Umnica [9.8K]
3 years ago
5

Given the following unbalanced chemical reaction: As + NaOH Na3AsO3 + H2 What would be the coefficient of the NaOH molecule in t

he balanced reaction?
a.1
b.2
c.4
d.6
Chemistry
1 answer:
barxatty [35]3 years ago
7 0

Answer: -

6

Explanation: -

The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2

We see there 3 sodium on the right side from Na3AsO3.

But there are only 1 sodium on the left from NaOH.

So we multiply NaOH by 3.

As + 3 NaOH -- > Na3AsO3 + H2

Now we see the number of Hydrogen on the left is 3.

But the number of hydrogens is 2 on the left.

So, we multiply to get both sides 6 hydrogen.

As + 6NaOH -- > Na3AsO3 + 3 H2

Rebalancing for Na,

As + 6NaOH -- > 2Na3AsO3 + 3 H2.

Finally balancing As,

2 As + 6 NaOH -- > 2Na3AsO3 + 3H2

The coefficient of the NaOH molecule in the balanced reaction is thus 6

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Explanation: both vessels are at the same temp and pressure (and the pressure is low and/or the temperature high).

6.7mol per 1.3L = 6.7/1.3 mol/L

so in 2.33L = 6.7*2.33/1.3 = 12 mol

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2 years ago
What is the pressure inside a 750 mL can of deodorant that starts at 15 degrees Celsius and 1.0 atm if the temperature is raised
Sonbull [250]

The answer is: the pressure inside a can of deodorant is 1.28 atm.

Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.  

p₁ = 1.0 atm.; initial pressure

T₁ = 15°C = 288.15 K; initial temperature.

T₂ = 95°C = 368.15 K, final temperature

p₂ = ?; final presure.

1.0 atm/288.15 K = p₂/368.15 K.  

1.0 atm · 368.15 K = 288.15 K · p₂.  

p₂ = 368.15 atm·K ÷ 288.15 K.  

p₂ = 1.28 atm.  

As the temperature goes up, the pressure also goes up and vice-versa.  

6 0
3 years ago
What is the volume in ML
BabaBlast [244]

Answer:1 mL

Explanation:

4 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
EleoNora [17]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

4 0
3 years ago
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