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Verizon [17]
3 years ago
14

Point How many grams of NaCl are present in 11.00 moles? ​

Chemistry
1 answer:
OlgaM077 [116]3 years ago
6 0

Answer:

642.8g

Explanation:

NaCl has a molar mass of 58.44g/mol, which means one mole of NaCl has a mass of 58.44g.

11 moles of NaCl will have a mass of 642.8g.

11 mol NaCl x 58.44g = 642.8g

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olga2289 [7]

Answer:

6 is the right answer I know cause I like science

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Determine how many grams of N2 are produced from the reaction of 8.37 g of H2O2 and 5.29 g of N2H4.
Dmitriy789 [7]
N₂H₄  +  2H₂O₂    →   N₂  +  4H₂O

mol = mass ÷ molar mass

If mass of hydrazine (N₂H₄)  = 5.29 g 
then mol of hydrazine           = 5.29 g ÷ ((14 ×2) + (1 × 4))
                                              = 0.165 mol

mole ratio of hydrazine to Nitogen is     1   :  1
  ∴ if moles of hydrazine = 0.165 mol
     then moles of nitrogen = 0.165 mol

Mass = mol × molar mass

Since mol of nitrogen (N₂)  = 0.165
then mass of hydrazine      = 0.165 × (14 × 2)
                                           = 4.62 g
5 0
3 years ago
Which of the
saul85 [17]

Answer:

During chemical reactions, matter is neither

created nor destroyed; it just changes form.

Explanation:

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Equator runs east and west all the way around the world,halfway between the south poles called the ______________and north poles
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3 0
3 years ago
2 NO + O22 NO2 is second order in NO and first order in O2. Complete the rate law for this reaction in the box below. Use the fo
riadik2000 [5.3K]

Answer : The value of rate of reaction is 1.35\times 10^{-8}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given chemical equation is:

2NO+O_2\rightarrow 2NO_2

Rate law expression for the reaction is:

\text{Rate}=k[NO]^a[O_2]^b

As per question,

a = order with respect to NO  = 2

b = order with respect to O_2 = 1

Thus, the rate law becomes:

\text{Rate}=k[NO]^2[O_2]^1

Now, calculating the value of rate of reaction by using the rate law expression.

Given :

k = rate constant = 9.87\times 10^3M^{-2}s^{-1}

[NO] = concentration of NO = 7.86\times 10^{-3}M

[O_2] = concentration of O_2= 2.21\times 10^{-3}M

Now put all the given values in the above expression, we get:

\text{Rate}=(9.87\times 10^3M^{-2}s^{-1})\times (7.86\times 10^{-3}M)^2\times (2.21\times 10^{-3}M)^1

\text{Rate}=1.35\times 10^{-8}Ms^{-1}

Hence, the value of rate of reaction is 1.35\times 10^{-8}Ms^{-1}

7 0
3 years ago
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