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Lelu [443]
2 years ago
7

Which word equation shows lithium oxide being formed from the reaction between oxygen and lithium? oxygen lithium oxide right ar

row. lithium lithium oxygen right arrow. lithium oxide oxygen lithium right arrow. lithium oxide lithium oxide right arrow. lithium oxygen
Chemistry
1 answer:
BlackZzzverrR [31]2 years ago
3 0

Answer:

lithium lithium (or 2lithium) + oxygen ->

Explanation:

Your welcome

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Why potasium more reactive than lithium and sodium
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So potassium is more reactive than lithium because the outer electron of a potassium atom is further from its nucleus than the outer electron of a lithium atom. Hope this answers the question. Have a nice day. Feel free to ask more questions.
5 0
3 years ago
When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?
Naddik [55]

Answer:

THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.

Explanation:

Mass = 3.0 kg = 3 * 1000 = 3000 g

Initial temperature = 10 C

Final temperature = 80 C

Change in temperature = 80 - 10 = 70 C

Specific heat of water = 4.18 J/g C

Heat needed = unknown

Heat is the amount of energy in joules needed to change a gram of water by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 3000 g * 4.18 J/g C * 70 C

Heat = 877 800 Joules

Heat = 877.8 kJ.

The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.

6 0
3 years ago
What are the answers to these questions
Virty [35]

Answer:

Explanation:

So basically it just asking you question about that surtain subject .

5 0
3 years ago
Which organism has a nerve net? A fish B hydra C flatworm D sponge
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5 0
3 years ago
Read 2 more answers
You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of
Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}

Amount of BaSO₄ from K₂SO₄ is;

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4} = X\times\frac{233.38 }{142.04} = 1.643·X

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4} = Y\times\frac{233.38 }{174.259 } = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0
3 years ago
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