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LekaFEV [45]
3 years ago
11

2. What are the names of the products when you mix an equal strength acid and alkali

Chemistry
1 answer:
dexar [7]3 years ago
8 0
I think the answer is water and hydrochloric acid
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How many moles of copper Il sulfide can be produced from 2.8 moles of sulfur? Cu + S --> CuS​
Kamila [148]

Answer: 2.8 moles of copper (Il) sulfide (CuS) will be produced from 2.8 moles of sulphur.

Explanation:

The balanced chemical reaction is:

Cu+S\rightarrow CuS  

According to stoichiometry :

1 mole of S produce =  1 mole of copper (Il) sulfide (CuS)

Thus 2.8 moles of S will produce=\frac{1}{1}\times 2.8=2.8moles of copper (Il) sulfide (CuS)

Thus 2.8 moles of copper (Il) sulfide (CuS) will be produced from 2.8 moles of sulphur.

5 0
3 years ago
For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
erastovalidia [21]

Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

Rate law 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.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

8 0
3 years ago
Please tell me the answer I will mark you brainliest
Shkiper50 [21]

Answer:

3

Explanation:

third answer might be right

8 0
3 years ago
Read 2 more answers
Why is the temperature different in different places in the U.S.
bagirrra123 [75]
Because of the equator 

5 0
3 years ago
Read 2 more answers
Determine the missing species:
Galina-37 [17]

Answer: 4. ^{0}_{-1}\textrm{e} and ^{0}_{1}\textrm{e}

Explanation:

a) The given reaction is ^{41}_{20}\textrm{Ca}+^{x}_{y}\textrm{X}\rightarrow ^{41}_{19}\textrm {K}

As the mass on both reactant and product side must be equal:

41+x=41

x=0

As the atomic number on both reactant and product side must be equal:

20+y=19

y=-1

^{41}_{20}\textrm{Ca}+^{0}_{-1}\textrm{e}\rightarrow ^{41}_{19}\textrm {K}

b) ^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{x}_{y}\textrm{X}

Total mass on reactant side = total mass on product side

15 =15 + x

x = 0

Total atomic number on reactant side = total atomic number on product side

8 = 7 + y

y = 1

^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{0}_{1}\textrm{e}

5 0
3 years ago
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