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soldier1979 [14.2K]
3 years ago
6

What are the rules for writing

Chemistry
1 answer:
Assoli18 [71]3 years ago
4 0
The rules of writing name is that they start with a big letter LIKE this and ends like this
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The cell of your body need energy to carry out life function. Which of the following body responses is a sign that your cells ne
Shkiper50 [21]
B is my pick that's what I think
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3 years ago
At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti
mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

\frac{R}{0.120}=\frac{1}{4}

R=0.03M/s

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0
3 years ago
As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis wit
jeka94

Answer:

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced  0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of  NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of  NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl  = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of  K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of  K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles =  0.0003663 moles

Mass of  K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%

7 0
3 years ago
What is the primary function of c6h12o6 in animals?
hichkok12 [17]

Answer:

Energy

Explanation:

C6H12O6 is glucose and the primary function of glucose is to provide energy.

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2 years ago
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