All categories

3 years ago

For the following hypothetical reaction:

Chemistry

1 answer:

nlexa [21]3 years ago

4 0

The number of B that will be needed to convert 5 moles of A into as many moles of C as possible is 20 moles

Explanation

3A + 12 B → 4 C

The moles of B is calculated using the mole ratio of A to B from equation above

that is the moles ratio of A:B = 3 : 12

Therefore the moles of B = 5 x12/3 = 20 moles

You might be interested in

Consider the following reaction between mercury(II) chloride and oxalate ion:

siniylev [52]

Answer : The reaction rate will be, $1.9\times 10^{-4}M/s$

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:

$2HgCl_2(aq)+C_2O_2^{4-}(aq)\rightarrow 2Cl^-(aq)+2CO_2(g)+HgCl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_2^{4-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_2^{4-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{k(0.164)^a(0.45)^b}{k(0.164)^a(0.15)^b}\\\\9=3^b\\(3)^2=3^b\\b=2$

Dividing 3 from 2, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{k(0.164)^a(0.45)^b}{k(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

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

Putting values in above rate law, we get:

$3.2\times 10^{-5}=k(0.164)^1(0.15)^2$

$k=8.7\times 10^{-3}M^{-2}s^{-1}$

Now we have to determine the reaction rate when the concentration of $HgCl_2$ is 0.135 M and that of $C_2O_2^{-4}$ is 0.40 M.

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

$\text{Rate}=(8.7\times 10^{-3})\times (0.135)^1\times (0.40)^2$

$\text{Rate}=1.9\times 10^{-4}M/s$

Therefore, the reaction rate will be, $1.9\times 10^{-4}M/s$

6 0

3 years ago

If an element has two isotopes, what is the atomic mass if one of the isotopes has a mass of 15.000 amu and makes up 5.000% of t

olasank [31]

The atomic mass of element is the weighted average atomic mass of the element with respect to the abundance of the isotopes of that element
atomic mass is the sum of the products of the mass of isotopes by their percentage abundance
atomic mass = 15.000 amu x 5.000 % + 16.000 amu x 95.000 %
= 0.7500 + 15.200
atomic mass of element is therefore 15.950

8 0

3 years ago

During which step in the pcr cycle do primers form bonds with a single-stranded template?.

shutvik [7]

Answer: The first steps

Explanation: science

5 0

2 years ago

Look at this fine human being. Hes 24 I kid you not.

ozzi

Answer:

um...

Explanation:

have a nice day and mark me brainliest! :)

Download docx

8 0

2 years ago

What percentage of strontium-90 remains in a sample after three half-lives have passed? 12.5% 15% 25% 60%?

Lyrx [107]

12.5% of strontium-90 would remain in a sample after three half-lives have passed. Half-life automatically means 50% of the original amount would remain.

4 0

3 years ago

Read 2 more answers