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klio [65]
2 years ago
6

If 38 gr of water are produced in reaction, how many mole are CO2 are produced.​

Chemistry
1 answer:
Nataly_w [17]2 years ago
6 0
0.86 moles of CO2 - i’m not sure if this is right but it should be something like that
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In which set do all elements tend to form anions in binary ionic compounds? li, na, k n, o, i c, s, pb k, fe, br?
pav-90 [236]

Atoms or molecule after gaining of electron possesses negative charge and is known as anion.

For the given sets:

  • Li, Na, K

The given elements are alkali metals and have tendency to lose electrons easily and form cations.

  • N, O, I

The given elements are non-metals and are electronegative. So, they gain electrons easily and form anion.

  • C, S, Pb

Carbon has tendency to form bond by sharing of electrons, Sulfur has tendency to gain electrons and form anion whereas Lead has tendency to lose electron.

  • K, Fe, Br

Potassium and Iron has tendency to lose electron and form cation whereas Bromine has tendency to gain electron to form anion.

Hence, from the given sets, all elements of set: N, O, I have tendency to form anions in binary ionic compounds.

8 0
3 years ago
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Which solution has the greatest osmolarity?
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Electric hidro because he has the highest osmolarity
4 0
3 years ago
How much heat is required to vaporize 31.5 gg of acetone (C3H6O)(C3H6O) at 25 ∘C∘C? The heat of vaporization for acetone at this
KiRa [710]

Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = \frac{mass}{molar mass}

Number of moles of acetone = \frac{31.5}{58.08}

Number of moles  of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

4 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
Likurg_2 [28]

Answer: 281 hours

Explanation:-

1 electron carry charge=1.6\times 10^{-19}C

1 mole of electrons contain=6.023\times 10^{23} electrons

Thus  1 mole of electrons carry charge=\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

96500\times 2=193000Coloumb of electricity deposits 1 mole or 63.5 g of copper

0.0635 kg of copper is deposited by 193000 Coloumb

11.5 kg of copper is deposited by=\frac{193000}{0.0635}\times 11.5=34952756 Coloumb

Q=I\times t

where Q= quantity of electricity in coloumbs  = 34952756 C

I = current in amperes = 34.5 A

t= time in seconds = ?

34952756 C=34.5A\times t

t=1013123sec=281hours

Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.

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the coast of California on the western side of the state has a voice climate wild Eastern side of the desert.what could cause th
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fires could cause a big difference in the weather of California

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