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tiny-mole [99]
3 years ago
12

An object becomes electrically charged when:

Chemistry
1 answer:
timurjin [86]3 years ago
7 0

Answer:

An object becomes electrically charged when <em>electrons</em><em> </em><em>are</em><em> </em><em>transferred</em><em> </em><em>to</em><em> </em><em>it</em>

Explanation:

<em>PLEASE</em><em> </em><em>DO MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>

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For about how long do scientists think wind and solar power can provide us with energy?
ivanzaharov [21]

Answer:

D

Explanation:

They are renewable energy sources

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What is the air temperature 8,850 m in the air?
evablogger [386]
The temperature is gonna be
5 0
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A 1.67-g sample of solid silver reacted in excess chlorine gas to give a2.21-g sample of pure solid Agcl.The heat given off in t
kotegsom [21]

<u>Given:</u>

Mass of Ag = 1.67 g

Mass of Cl = 2.21 g

Heat evolved = 1.96 kJ

<u>To determine:</u>

The enthalpy of formation of AgCl(s)

<u>Explanation:</u>

The reaction is:

2Ag(s) + Cl2(g) → 2AgCl(s)

Calculate the moles of Ag and Cl from the given masses

Atomic mass of Ag = 108 g/mol

# moles of Ag = 1.67/108 = 0.0155 moles

Atomic mass of Cl = 35 g/mol

# moles of Cl = 2.21/35 = 0.0631 moles

Since moles of Ag << moles of Cl, silver is the limiting reagent.

Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles

Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol

Ans: Formation enthalpy = 126.5 kJ/mol


6 0
3 years ago
Read 2 more answers
Need help on 17,19,20 please.
pentagon [3]

Answer:

c, maybe d, and I think b.

Explanation:

Im sorry if wrong

5 0
3 years ago
Carbon dioxide and water react to form methanol and oxygen, like this:
fredd [130]

Answer:

24x10³

Explanation:

2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)

The equilibrium constant for this reaction is:

Kc = \frac{[O_2]^3}{[CO_2]^2[H_2O]^4}

The expression of [CH₃OH] is left out as it is a pure liquid.

Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:

  • CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
  • H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
  • O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂

Then we calculate the concentrations:

  • [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
  • [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
  • [O₂] = 0.0875 mol / 7.5 L = 0.0117 M

Finally we <u>calculate Kc</u>:

  • Kc = \frac{0.0117^3}{0.0099^2*0.0285^4} = 24x10³

3 0
3 years ago
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