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2 years ago

Which is a factor in determining the average atomic mass of an element

1 answer:

4 0

The mass numbers of the different isotopes of that element are averaged according to their respective abundances in nature.

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Calculate the net force 175 N 1125 N

Ket [755]

Answer:

<h3>_____________&_________</h3><h2>Net force = sum of all forces with directions</h2><h2>so.....</h2><h2>NET FORCE = 125+(-75)</h2><h2> = 50 N</h2><h2> in the direction of yellow </h2><h2> t-shirt man</h2><h2>------------------------------------</h2>

4 0

3 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

Is flat soda mixture or pure substance?

Nesterboy [21]

The answer should be Pure substance

5 0

2 years ago

What is the volume of 0.120 g of C2H2F4 vapor at 0.970 atm and 22.4°C? Answer in units of L.

ludmilkaskok [199]

Answer:

V= 0.031L

Explanation:

P= 0.97atm, V= ?, n= 0.12/98 =0.00122mol, R= 0.082, T= 22.4+273= 295.4

Applying

PV=nRT

0.970×V = 0.00122×0.082×295.4

Simplify the above equation

V= 0.031L

6 0

2 years ago

A sample of a vapor occupies a volume of 500 mL at 65°C. If pressure remains constant, what is the volume of the gas at standard

DedPeter [7]

Answer:

403 mL

Explanation:

First, I will assume that the mole is 1, because you are not specifing this.

Now, with the innitial data, we need to get the pressure:

T = 65+273 = 338 K

V = 500 / 1000 = 0.5 L

Now if:

PV = nRT

Then:

P = nRT/V and V = nRT/P

Let's calculate the P:

P = 1 * 0.082 * 338 / 0.5 = 55.432 atm

The standard temperature is 0° C or 273 K so, the volume is:

V = 1 * 0.082 * 273 / 55.432

V = 0.40384 L or simply 403.84 mL

8 0

3 years ago