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3 years ago

How many oxygen atoms are present in 2.5 mol

Chemistry

1 answer:

gladu [14]3 years ago

3 0

15 X 10 ^23 atoms of O2

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9474 millimeters to centimeters using scientific notation showing work

puteri [66]

Answer:

9.474 x 10^2

Explanation:

ok. first you have to get the value in the required unit so 9474mm/(10mm/cm) = 947.4 so scientific notation states that the number must be raised to any power of an integer and the value of the number being raised must be less than than 10 and more than or equal to 1

so it must have one digit in front so.. 947.4 becomes 9.474 and because you move 2 places to the left, ur power is positive 2

and proof 10^2 is 100 so multiply 9.474 by 100 and u will get 947.4 cm which is also 9474 mm

8 0

2 years ago

The chemical formula for beryllium oxide is BeO.A chemist determined by measurements that 0.045 moles of beryllium participated

blondinia [14]

Answer: 0.405g

Explanation:

Molar Mass of Be = 9g/mol

Number of mole of Be = 0.045mol

Mass conc. Of Be = 0.045 x 9 = 0.405g

8 0

3 years ago

Stock naming please help

andre [41]

Answer:

Can you tell me the question in the comments on this answer or like how you do it then ill answer you in the comments under this answer

8 0

2 years ago

Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$

But this is just a supposition.

Regards.

4 0

2 years ago

_____ NaPO4+_____KOH

Artist 52 [7]

NaPO4 + KOH -> KPO4 + NaOH
already balance

6 0

2 years ago