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djverab [1.8K]
3 years ago
9

A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3 . calculate the density of gold.

Chemistry
1 answer:
Fantom [35]3 years ago
6 0

Answer:

            19.29 g.cm⁻³

Solution:

Data Given:

                             Mass  =  301 g

                             Volume  =  15.6 cm³

Formula Used:

                             Density  =  Mass ÷ Volume

Putting values,

                             Density  =  301 g ÷ 15.6 cm³

                            Density  =  19.29 g.cm⁻³

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viva [34]

Answer:

The second and third option are isotopes of Bromine

Z=  35 A = 79

A = 79  N= 44

Explanation:

Step 1: What are isotopes ?

⇒ Elements  with same atomic number (Z) ( this means the same number of electrons and protons) but a different number of neutrons (N)

The atomic number (visible on the periodic table) is the number of protons.

The atomic mass is  the sum of the protons (Z) and neutrons (N), and is showed as 'A'. So A = Z+N

If we look at the periodic table, we can see that the atomic number of bromine (Z) = 35. This means the amount of protons = 35. Since isotopes have the same amount of protons, all isotopes of bromine, have 35 protons.

1) Z = 79, A = 196

Z = protons = electrons .So this element has 79 protons, as well as 79 electrons.

196 = 79 + N ⇒ N = 196 - 79 = 117 neutrons

⇒Since it doesn't have 35 protons, it isn't an isotope of Bromine, but of Gold (Au): 196Au

2) Z = 35 , A = 79

Z = protons = electrons .So this element has 35 protons, as well as 35 electrons.

79 =35 + N ⇒ N = 79 - 35= 44 neutrons

⇒Since it does have 35 protons, it is an isotope of Bromine. This isotope has 44 neutrons

⇒ This is 79Br, which is a stable isotope of Bromine.

3) A=79, N = 44

Z = A - N ⇒ Z = 79 - 44 = 35

Z = protons = electrons .So this element has 35 protons, as well as 35 electrons. It also has 44 neutrons.

⇒Since it does have 35 protons, it is an isotope of Bromine, with 44 neutrons: 79Br

4) Z = 44, N = 44

Z = protons = electrons .So this element has 44 protons, as well as 44 electrons.

A = 44 + 44 ⇒ A = 88  

⇒Since it doesn't have 35 protons, it isn't an isotope of Bromine, but of Ruthenium (Ru) : 88Ru

⇔So the <u>second</u> and<u> third</u> option are isotopes of Bromine

6 0
3 years ago
Consider the following multistep reaction:
chubhunter [2.5K]

The question is incomplete, here is the complete question:

Consider the following multistep reaction:

C+D⇌CD (fast)

CD+D→CD₂ (slow)

CD₂+D→CD₃ (fast)

C+3D→CD₃

Based on this mechanism, determine the rate law for the overall reaction.

<u>Answer:</u> The rate law for the reaction is \text{Rate}=k'[C][D]^2

<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

C+3D\rightarrow CD_3

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  C+D\rightleftharpoons CD;\text{ (fast)}

<u>Step 2:</u>  CD+D\rightarrow CD_2;\text{(slow)}

<u>Step 3:</u>  CD_2+D\rightarrow CD_3;\text{(fast)}

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

\text{Rate}=k[CD][D]           ......(1)

As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for CD from step 1, we get:

K=\frac{[CD]}{[C][D]}

[CD]=K[C][D]

Putting the value of [CD] in equation 1, we get:

\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2  

Hence, the rate law for the reaction is \text{Rate}=k'[C][D]^2

5 0
3 years ago
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