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Mariulka [41]
3 years ago
15

How many Ni atoms are in 3.6 mol of Ni

Chemistry
1 answer:
Sliva [168]3 years ago
6 0
I believe that the answer is 1.8^24 of Ni atoms in 3.6 mol of Ni.

Hope this helps. :)

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A gas sample at stp contains 1.15 g oxygen gas and 1.55 g nitrogen gas.what is the volume of the gas sample?
ss7ja [257]
O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol

4 0
3 years ago
Read 2 more answers
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
Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 1.731.
Amanda [17]

Answer:

The answer to your question is letter c) 6.09 g of sodium and 9.38 g of chlorine.

Explanation:

This problem is solve using rule of three

We know that the proportion Sodium to Chloride is 1 to 1 in sodium chloride, so we have to look for this proportion in the options

AM Sodium = 23 g

AM Chlorine = 35.5 g

                 Sodium                                     Chlorine

           23 g ---------------- 1 mol              35.5 g -------------- 1 mol

       1713.73g -------------    x               2666.6 g -------------    x

          x = 1713.73/23 = 74.51                     x = 2666.6/35.5 = 75.12

    These values are very similar, we have to look for the proportion in the options

a)      6.09g of sodium = 0.26 mol

      4,87 g of chlorine = 0.14 mol           These numbers are not very similar

b) We have 0.26 mol of Na

                  0.037 mol of Cl                      This is not the answer

c) We have 0.26 mol of Na

                  0.26 mol of Cl                     These numbers are the same, the proportion is 1:1, this is the answer

d) We have 0.26 mol of Na

                   0.36 mol of Cl                    This is not the answer

8 0
3 years ago
concentrated phosphoric acid is90% H3PO4 by mass and the remaining mass is water. The molarity of H3PO4 is 12.2M at temperature
blondinia [14]

Answer:

82.0 mL

Explanation:

Step 1: Given data

  • Concentration of concentrated acid (C₁): 12.2 M
  • Volume of concentrated acid (V₁): ?
  • Concentration of dilute acid (C₂): 1.00 M
  • Volume of dilute acid (V₂): 1.00 L

Step 2: Calculate the required volume of the concentrated acid

We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL

4 0
2 years ago
Is dew forming on a leaf a physical or chemical change?
solong [7]
No, the formation of dew is condensation, which is a physical change.
5 0
3 years ago
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