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Temka [501]
2 years ago
7

Calculate the number of atoms in 3 moles of carbon atoms​

Chemistry
1 answer:
creativ13 [48]2 years ago
6 0

Answer:

1,806,000,000,000,000,000,000,000 atoms.

Explanation:

1 mol of anything is 6.022*10^23 units

1 mol carbon atoms = 6.022*10^23 carbon atoms

3 mol carbon atoms = 3*6.022*10^23 = 1.8*10^24 atoms

Correctly - your answer should have only 1 significant digit: 2 * 10^24 atoms.

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Ammonia can be produced in the laboratory by heating ammonium chloride
AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl   +  Ca(OH)₂     →    CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

                      NH₄Cl          :          NH₃

                          2              :           2

                         0.17          :          0.17

                   Ca(OH)₂         :          NH₃

                        1                :           2

                    0.10              :          2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

6 0
3 years ago
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
4 years ago
Which compounds are classified as Arrhenius acids?
Olenka [21]

Answer:The correct answer is option 4.

Explanation:

Arrhenius acids are those compounds which gives H^+ ions when dissolved in their aqueous solution.

HA(aq)\rightarrow H^++A^-

Arrhenius bases are those compounds which gives OH^- ions when dissolved in their aqueous solution.

BOH(aq)\rightarrow OH^-+B^+

HBr \& H_2SO_4 are Arrhenius acids because they form H^+ions in their respective aqueous solution.

HBr(aq)\rightarrow H^++Br^-

H_2SO_4(aq)\rightarrow 2H^++SO_{4}^{2-}

Hence, the correct answer is option 4.

6 0
3 years ago
Read 2 more answers
A solid that forms and separates from a liquid mixture is a.......
inna [77]

A solid that forms and separates from a liquid mixture is a chemical change.

4 0
4 years ago
A buffer is formed by mixing 100 ml of 0.1 m nh4cl with 50 ml of 0.3 m nh3. to this solution 13 ml of 0.1 m naoh is added. what
Hoochie [10]
I had the same question last week and my answer was C. Try that.
8 0
4 years ago
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