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ArbitrLikvidat [17]
3 years ago
13

Apply the Law of Conservation of Mass to the following problem: During a combustion reaction, 12.2 grams of methane reacts with

14 g of oxygen. The reaction produces carbon dioxide and water. If 20 grams of water are produced, how many grams of carbon dioxide are produced?
Chemistry
1 answer:
Ratling [72]3 years ago
5 0

Answer:

5.8 g of carbon dioxide are produced

Explanation:

The Law of Conservation of Mass states that the mass of the reactants must equal the mass of the products in all chemical reactions.

This is the chemical reaction (combustion)

 CH₄     +   2O₂    →    CO₂   +    2H₂O

12.2 g        14 g             x             20g

Mass in reactants = 12.2 g + 14 g = 26.2 g

Mass in products =  x + 20 g

26.2 g = x + 20g

26.2 g - 20g = x

5.8 g = x

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what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?
MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now  F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻  +  H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF,  the [ OH ⁻] from the given pH, and finally the mass needed to produce that  OH⁻ concentration.  

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is  6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14  ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

                                   F⁻                    HF                        OH⁻

Equilibrium                 X                  2.51 x 10⁻⁶            2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X  =  1.5 x 10⁻¹¹     ⇒  X =  (2.51 x 10⁻⁶)²  / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L  *  0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF *  42.0 g/mol NaF = 6.03 g NaF

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The half-life of sr-90 is 28 years. after 56 years of decay only 0. 40 g of a sample remains. what was the mass of the original
Svetradugi [14.3K]

Half-life is the time taken for the concentration of the substance to reduce by 50%. The original sample of strontium had a mass of 1.6 gms. Thus, option d is correct.

<h3>What is half-life?</h3>

The half-life of any radioactive substance is the time period at which the concentration will get reduced to half the initial amount. The initial mass of Sr-90 is calculated as,

N(t) = N_{0} (\dfrac{1}{2})^{ \frac{t }{t 1/2}}

Given,

Quantity of the remaining substance N (t) = 0.40 gm

Initial radioactive substance quantity N_{0} =?

Time duration (t) = 56 years

Half-life = 28 years

Substituting values above:

\begin{aligned} 0.40 &= N_{0} (\dfrac{1}{2}) ^{{\frac{56}{28}}\\\\0.40 &= N_{0} (\dfrac{1}{2})^{2}\end{aligned}

= 1.6 gm

Therefore, option d. the initial mass of Sr is 1.6 gm.

Learn more about half-life here:

brainly.com/question/16145921

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