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alexgriva [62]
3 years ago
12

Butter has a specific gravity of 0.86. What is the mass of 2.15 L of butter?

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Answer:

The mass of butter is 1849 g.

Explanation:

Given data:

Specific gravity of butter = 0.86

Volume = 2.15 L

Mass = ?

Solution:

Specific gravity of a substance is the ratio of density of substance divided by the density of water.

Thus the density of butter is 0.86 g/mL

Formula:

Conversion from L to mL

2.15 /1000 = 2150 mL

d = m/v

0.86 g/mL = m/2150 mL

m =  1849 g

Thus the mass of butter is 1849 g.

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A carbon skeleton is covalently bonded to both an amino group and a carboxyl group. When placed in water it ____
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Answer:

The correct option is: (D) would function as both an acid and a base

Explanation:

A carbon skeleton bonded to a amino group as well as a carboxyl group, will behave as an acid in basic medium and base in acidic medium. This is because the carboxyl group present in the compound will release a proton in basic medium and the amino group will accept a proton in the acidic medium.

<u>Therefore, a carbon skeleton which is covalently bonded to a carboxyl and amino group will behave as both acid and base.</u>

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Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi
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Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 2.9^0C

Explanation:

Elevation in boiling point:

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of pure carbon disulfide= 46.2^oC

k_b = boiling point constant  =2.34^0Ckg/mol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 35.0 g

w_1 = mass of solvent (carbon disulphide) = 397.7 g

M_2 = molar mass of solute = 70.0 g/mol

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

(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}

T_b=49.1^0C

Therefore, the change in boiling point is (49.1-46.2)^oC=2.9^0C

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A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2
Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

<u>Step 1: </u>Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

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<u>Step 2</u>: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

<u>Step 3:</u> Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

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<u>Step 4: </u>Calculate molarity of H2C2O4

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Molarity H2C2O4 = 0.00428 moles / 0.250 L

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2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

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