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Norma-Jean [14]
3 years ago
12

How many moles in 28 grams of UO2?

Chemistry
2 answers:
Leya [2.2K]3 years ago
8 0

Answer:

There are 0.135moles of UO2 in 28 grams of UO2

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 mole UO2, or 207.03grams.

This is the molar mass of UO2, calculated from the sum of the molar masses of uranium and oxygen.

That is, molar mass of UO2= molar mass of U + (molar mass of O)2

=238.03g/mol + (16g/mol)2

= 238.03g/mol + 32g/mol= 270.03g/mol

Molar mass of UO2= 270.03g/mol

This implies that 1 mole of UO2 has a mass of 207.03 grams

Number of moles of UO2 = mass of UO2/ molar mass of UO2

Mass of UO2= 28 grams

Number of moles of UO2= 28 g/ 207.03gmol-1

Number of moles =0.1352 moles of UO2

Therefore, there are 0.135 moles of UO2 in 28 grams of UO2.

Lina20 [59]3 years ago
3 0
0.22 moles.

28 divided by 124 =0.22
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4 years ago
The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.
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Answer:

c    43.38 g

Explanation:

The reaction  between MnO2 and HCl can be represented by the following balanced equation:

MnO2 + HCl ---> Cl2 + MnCl2 + H2O

From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.  

For the fact that HCl  gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess,  while the limiting reagent should be MnO2 .  

Thus, the theoretical yield of Cl2 will be  60.25 g.

By definition, the percentage yield is given by

% Yield = (Actual Yield) / (Theoretical Yield),  

This can be simplified to

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Plugging in the given values we have

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5 0
3 years ago
g Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at
Alborosie

Answer :  The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

where,

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R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

Q  = reaction quotient

[A]_{inside} = concentration inside the cell

[A]_{outside} = concentration outside the cell

Now put all the given values in the above formula, we get:

-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

\frac{[A]_{inside}}{[A]_{outside}}=296.2

Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

3 0
3 years ago
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