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s2008m [1.1K]
2 years ago
6

45.0 g of Ca(NO3)2 are used to create a 1.3 M solution. What is the volume of the solution

Chemistry
1 answer:
Stells [14]2 years ago
6 0
As we know that Molarity is given as,

                                       M = moles / V 
Solving for V,
                                        V = moles / M ------------------(1)
Also, moles is equal to,
                                       moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
                                       V = (mass / M.mass) / M
Putting values,
                                       V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³ 
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At STP, a 50-gram sample of H20(I) and a 100-gram sample of H20(I) have
Katyanochek1 [597]

Answer:

(1) the same chemical properties .

Explanation:

Hello!

In this case, among the options:

(1) the same chemical properties

(2) the same volume

(3) different temperatures

(4) different empirical formulas

We can see that they have the same chemical properties as they at the same conditions, same type of bond (polar), molecular geometry, bond angle (104.5 °) and so on. Nevertheless, at STP (1 atm and 273.15 K) they do not have the same volume since the larger the mass, the larger the volume, they have the same temperature and the both of them are H₂O.

It means that the answer is (1) the same chemical properties .

Best regards.

3 0
2 years ago
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Janine is trying to separate some ethanol from water. Which method should she use does anyone know this
GuDViN [60]

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Simple distillation is a separation method in which two liquids can be separated from each other based on differences in boiling point.

Since water has a higher boiling point than ethanol, ethanol is collected first as the distillation proceeds.

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2 years ago
Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,00
AnnZ [28]

Answer:

Hello some parts of your question is missing

Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.

Answers:

a) 46.666.66 g/mol

b) 20930.23 g/mol

c)43333.33 g/mol

Explanation:

A)The equal number of molecules of each sample can be calculated using  Mn = \frac{n(M1 + M2 + M3)}{3n}

because for the number of molecules to be equal : n1 = n2 = n3 = n

Mn = 46666.66 g/mol

B ) To calculate the equal masses of each sample

we apply this equation

Mn = \frac{W1 + W2 +W3}{\frac{W1}{M1} +\frac{W2}{M2}+ \frac{W3}{M3}  }

ATTACHED BELOW IS THE REMAINING PART OF THE SOLUTION

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