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2 years ago

Helpppppo meeeeee plzzzzzzzzz

Chemistry

1 answer:

antoniya [11.8K]2 years ago

6 0

For 2 it’s 2ft 10in
For 5 it’s 3ft 6in
You: put how tall you are
Adult: you can either measure an adult or use this 5 ft 10in

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Balance the following equations:

Brilliant_brown [7]

Answer:

1) 0 C2H4O2 + 0 O2 -> 0 CO2 + 0 H2O (balanced)

2) V2O5 + CaS -> CaO + V2S5

just additional info: V2O5 is divanadium pentaoxide

LHS (Left hand side)

V: 2

O: 5

Ca: 1

S: 1 x 5 [to balance with the right hand side of the equation]

RHS (Right hand side)

V: 2

O: 1 x 5 [to balance with the left hand side of the equation]

Ca: 1

S: 5

When you balance any elements, you have to balance the whole chemical compound.

Thus,

V2O5 + 5 CaS -> 5 CaO + V2S5

LHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

RHS CHECK:

V: 2

O: 5

Ca: 5

S: 5

3) S8 + O2 -> SO2

LHS:

S: 8

O: 2

RHS:

S: 1 x 8 [to balance with LHS]

O: 2

When you balance any elements, you have to balance the whole chemical compound.

S8 + O2 -> 8 SO2

When we add 8 to the RHS, it gives us 8S, 16 O.

In order to balance that into the RHS, I need to multiply the O2 by 8, which will give 8(O2) = 16 O particles.

Therefore, S8 + 8 O2 -> 8 SO2 is the final answer for (3).

6 0

3 years ago

The energy required to completely separate the ions from a solid is

Misha Larkins [42]

The answer is (B.) lattice energy

5 0

3 years ago

A 13.5 g sample of an unknown gas occupies 5.10 L at 149.83 kPa and 301 K. What is the molar mass of the gas ?

alisha [4.7K]

Answer:

The molar mass of the gas is 44.19 g/mol

Explanation:

Amount of sample of gas = m = 13.5 g

Volume occupied by the gas = V = 5.10 L

Pressure of the gas = P = 149.83 KPa

1 KPa = 0.00986 atm

P = $149.83 \textrm{ KPa} \times 0.00986 \textrm{ atm/KPa} = 1.48 \textrm{ atm}$

Assuming M g/mol to be the molar mass of the gas

Assuming the gas is behaving as an ideal gas

$\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\1.48 \textrm{ atm}\times 5.10 \textrm{ L} = \displaystyle \frac{13.5 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 301\textrm{K} \\M = 44.19 \textrm{ g/mol}$