Magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide as shown.

. Given the reaction 2HgO → 2Hg + O2 , how many moles of elemental mercury will be obtained by the decomposition of 1 mole of Hg

AveGali [126]

Answer:

one mole of HgO will give one mole of Hg.

Explanation:

Given data:

Moles of HgO = 1 mol

Moles of Hg = ?

Solution:

Chemical equation:

2HgO → 2Hg + O₂

Now we will compare the moles of Hg with HgO from balance chemical equation.

HgO : Hg

2 : 2

1 : 2/2×1 = 1 mol

So, one mole of HgO will give one mole of Hg.

6 0

3 years ago

What is the atomic mass of a carbon isotope that has 6 protons and 7 neutrons?

DedPeter [7]

The atomic mass of a carbon isotope that has 6 protons and 7 neutrons is 13

Explanation:

Remember that whilst the atomic number represents the number of protons in an atom, the mass number represents the summation of protons and neutrons particles in the atomic nuclei. Therefore, in this case, the carbon will have a mass number of;

6 + 7 = 13

Isotopes of an element usually have the same atomic number but different mass numbers -because they have slightly different numbers of neutrons. An example is isotopes of Carbon; C-14 and C-12

7 0

3 years ago

Read 2 more answers

Identify two ways that the rate of a chemical reaction could be increased

Svetlanka [38]

Increase the Surface Area of the Reactants. Increasing the surface area of the reactants increases the rate of the reaction. More surface area means more collisions of the reactant molecules and an increased rate of the reaction. This occurs when reactants are made to react in powdered form

7 0

4 years ago

Do all multicellular organisms follow the same pattern of organization?

natta225 [31]

So so sorry if i'm wrong but i'm pretty sure no

3 0

4 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago