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3 years ago

What do you call the ability of a gas to completely spread out and mix with other substances?

Chemistry

2 answers:

olganol [36]3 years ago

6 0

The ability for gas to mix with other substances is diffusion.

ASHA 777 [7]3 years ago

4 0

Explanation:

When a gas is able to spread out and mix completely with other substances then this property of gases is known as diffusion.

It is known that molecules of a gas are held together by weak Vander waal forces hence, they move rapidly from one place to another and easily spread into the atmosphere.

Also, molecules of a gas spread randomly and does not follow any particular order to move.

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How can you tell the mass was conserved in a chemical reaction?

KengaRu [80]

Answer:

$\boxed {\boxed {\sf Mass \ of \ products \ = \ mass \ of \ reactants}}$

Explanation:

The Law of Conservation of Mass states that mass cannot be created or destroyed.

In a chemical reaction, there are products and reactants. The reactants react and produce the products.

Since mass can't be created or destroyed, the mass of the products must be equal to the mass of the reactants if mass was conserved in a chemical reactions.

6 0

3 years ago

Balance this equation- Fe + O2 Fe3 O4

kherson [118]

3Fe + 2O2 —- Fe3 + O4

7 0

2 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago

Help me help me please

Vadim26 [7]

Answer:

C. wind

Explanation:

4 0

3 years ago

Read 2 more answers

What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?

insens350 [35]

First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm

6 0

3 years ago