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1 year ago

A student needed exactly 45.3ml of a solution. what piece of glassware should that student use? justify your choice.

1 answer:

7 0

The student should use the graduated cylinder. A graduated is the most common laboratory glassware when measuring volumes. It has calibrations by 1, 0.5 or 0.1 depending on the maximum volume. You have to make sure though, that you measure the volume by looking at the lower meniscus of the liquid at eye level.

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Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

1 year ago

In each pair of statements, choose the example that is more reliable.

TEA [102]

Answer:

"statement 2" for the first pair and "statement 1" for the second pair

Explanation:

3 0

1 year ago

Read 2 more answers

A compound that contains only carbon, hydrogen, and oxygen is 48.64% C and 8.16% H by mass. What is the empirical formula of thi

TiliK225 [7]

Answer:

The empirical formula of this substance is:

$C_3H_6O_2$

Explanation:

To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:

- C: 12.01 g/mol

- H: 1.00 g/mol

- O: 15.99 g/mol

With the information in this exercise we can suppose in 100 g of the substance we have:

C: 48.64 g

H: 8.16 g

O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)

Now, we need to divide these grams by the molecular weight:

$C=\frac{48.64g}{12.01 g/mol} =4.05 mol\\H=\frac{8.16g}{1.00g/mol}= 8.16 mol\\O=\frac{43.2g}{15.99 g/mol} = 2.70 mol$

We need to divide these results by the minor result, in this case O=2.70 mol

$C=\frac{4.05mol}{2.70mol}= 1.5\\H= \frac{8.16mol}{2.70 mol} = 3.02 \\O=\frac{2.70mol}{2.70mol} = 1$

We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:

$C= 1.5*2=3\\H= 3.02*2= 6.04 \\O= 1*2=2$

This numbers are very close to integer numbers, so we can find the empirical formula as subscripts in the chemical formula:

$C_3H_6O_2$

6 0

1 year ago

A silver sphere has a mass of 5.492 g and a diameter of 10.0 mm. What is the density of silver metal in grams per cubic centimet

Blizzard [7]

Answer:

Explanat $\rho=10.5\frac{g}{cm^3}$ ion:

Hello,

In this case, considering the given diameter which is related to a radius of 5.0 mm and the formula for the calculation of the volume of the sphere, its volume in cubic centimeters (5.00 mm = 0.5 cm) is then:

$V=\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*(0.5cm)^3\\\\V=0.524cm^3$

In such a way, the density turns out:

$\rho =\frac{m}{V} =\frac{5.492g}{0.048m^3} \\\\\rho=10.5\frac{g}{cm^3}$

Best regards.

8 0

1 year ago

Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord

Marina CMI [18]

Answer:

$M_{acid}=0.563M$

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the molarity of the acid to obtain:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M$

Regards!

5 0

1 year ago