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2 years ago

A student was directed to convert the mass of a boulder from kg to pounds.

Chemistry

1 answer:

Alina [70]2 years ago

5 0

Answer:

Incorrect

Explanation:

Given that,

The mass of a boulder, m = 22 kg

We need to find the weight of the student in pounds.

We know that,

1 pound = 454 g

1 g = (1/454 pounds)

22 kg = 0.022 grams

So,

0.022 grams = $\dfrac{0.022}{454}\ pounds$

= 0.000048 pounds

The student says the weight is 0.048 pounds. It means he was incorrect.

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50 POINTS!!! DUMB ANSWERS WILL BE DELETED

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3 years ago

The teacher in item 10 also needs to order plastic tubing. If each of the 60 students need 750 mm of tubing, what length of tubi

Alex777 [14]

45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.

a) Find the length in millimetres

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3 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$

Hence, the molar mass of the insulin is 6087.2 g/mol

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3 years ago

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NikAS [45]

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2 years ago

Read 2 more answers

The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r

Vinil7 [7]

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

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2 years ago