For a constant force, when the surface area is tripled the

The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 17

yawa3891 [41]

Answer:

a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

b. 0.957 g

Explanation:

Step 1: Write the balanced equation

2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

Step 2: Convert 130.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15

K = 130.0°C + 273.15

K = 403.2 K

Step 3: Calculate the moles of O₂

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K

n = 2.21 × 10⁻³ mol

Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂

The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.

Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO

The molar mass of HgO is 216.59 g/mol.

4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g

5 0

3 years ago

Help please

Mars2501 [29]

Answer:

electron cloud charge of manganese=-5

Explanation:

atomic number=number of elecron +neutron number

DATA

Neutron NO.=30

atomic NO.=25

e=?

A=e+n

25=e+30

e=25-30

e= -5

6 0

3 years ago

Calculate the molality of a solution prepared by dissolving 175 g of KNO3 in 750 g of water. (round to nearest hundreth)

Wittaler [7]

We will get the molality from this formula:
Molality = no.of moles of solute / Kg of solvent

So first we need the no.of moles of KNO3 = the mass of KNO3 / molar mass of KNO3
no.of moles of KNO3 = 175 / 101.01 = 1.73 mol
By substitution in the molality formula:
∴ molality = 1.73 / (750/1000) = 2.3 Molal

6 0

3 years ago

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Balance the following reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coeffici

irina1246 [14]

Answer:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Explanation:

Step 1: Write the unbalanced chemical equation,

C₄H₁₀ + O₂ → CO₂ + H₂O

Step 2: Balance Carbon Atoms;

As there are 4 carbon atoms on left hand side and 1 carbon atoms on right hand site therefore, to balance them multiply CO₂ on right hand side by 4 i.e.

C₄H₁₀ + O₂ → 4 CO₂ + H₂O

Step 3: Balance Hydrogen Atoms;

There are 10 hydrogen atoms on left hand side and 2 hydrogen atom on right hand site therefore, to balance them multiply H₂O on left hand side by 5 i.e.

C₄H₁₀ + O₂ → 4 CO₂ + 5 H₂O

Step 4: Balance Oxygen Atoms:

Now there are 2 oxygen atoms in reactant side and 13 in product side. So, multiply O₂ by 6.5 i.e.

C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

Step 5: Remove fraction coefficients as,

Multiply whole equation by 2 to get rid of fractions i.e.

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

7 0

3 years ago

As the atomic numbers of the elements in column IIA increase, the size of the atoms generally

joja [24]

Answer:

Increase

Explanation:

7 0

3 years ago

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