The isotope 65Cu2+ has how many electrons and how many protons?

In a laboratory experiment, students synthesized a new compound and found that when grams of the compound were dissolved to make

marshall27 [118]

The question is incomplete, the complete question is;

In a laboratory experiment, students synthesized a new compound and found that when 12.23 grams of the compound were dissolved to make 228.1 mL of a benzene solution, the osmotic pressure generated was 4.55 atm at 298 K. The compound was also found to be non-volatile and a non-electrolyte. What is the molecular weight they determined for this compound?

Answer:

287.76 g/mol

Explanation:

From;

π = M R T

M = molarity

R= gas constant

T = temperature

number of moles = π * volume/RT

number of moles = 4.55 * 228.1/1000/0.082 * 298

number of moles = 1.037855/24.436

number of moles = 0.0425 moles

Molar mass = mass/number of moles

Molar mass = 12.23 grams/0.0425 moles

Molar mass = 287.76 g/mol

4 0

3 years ago

A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate

inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w = (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 / 1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0

3 years ago

24 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Katena32 [7]

The answer is B my guy.

4 0

3 years ago

Read 2 more answers

how many grams of calcium oxide will be produced in a closed vessel containing 20.0 kg of calcium and 20.0 kg of oxygen gas if t

sergejj [24]

Answer:

27984.1311 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For calcium :-

Mass of calcium = 20.0 kg = 20000 g ( 1 kg = 1000 g )

Molar mass of calcium = 40.078 g/mol

Thus,

$Moles= \frac{20000\ g}{40.078\ g/mol}$

$Moles\ of\ calcium= 499.0269\ mol$

For oxygen gas :-

Mass of oxygen gas = 20.0 kg = 20000 g ( 1 kg = 1000 g )

Molar mass of oxygen gas = 31.999 g/mol

Thus,

$Moles= \frac{20000\ g}{31.999\ g/mol}$

$Moles\ of\ oxygen\ gas= 625.0195\ mol$

According to the given reaction:

$2Ca+O_2\rightarrow 2CaO$

2 moles of calcium reacts with 1 moles of oxygen gas

Also,

1 mole of calcium react with 1/2 mole of oxygen gas

So,

499.0269 mole of calcium react with $\frac{1}{2}\times 499.0269$ mole of oxygen gas

Moles of oxygen gas = 249.5135 moles

Available moles of oxygen gas = 625.0195 moles

Limiting reagent is the one which is present in small amount. Thus, calcium is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

2 moles of calcium produces 2 moles of calcium oxide

So,

1 mole of calcium produces 1 mole of calcium oxide

Also,

499.0269 mole of calcium produces 499.0269 mole of calcium oxide

Moles of calcium oxide = 499.0269 moles

Molar mass of calcium oxide = 56.0774 g/mol

Mass of calcium oxide = Moles × Molar mass = 499.0269 × 56.0774 g = 27984.1311 g

27984.1311 g of calcium oxide will be produced.

8 0

4 years ago

Shaun's doctor has recommended that he consume 200 calories of protein a day. Shaun consumed 176 calories of protein in one day.

evablogger [386]

Answer: Shaun must consume 6 more g of protein to reach the required amount.

Explanation:

Given : Recommended amount of protein = 200 calories

Now 4 calories = 1 gram of protein

Thus 200 calories = $\frac{1}{4}\times 200=50$ g of protein.

Calories consumed per day by Shaun = 176 calories

176 calories = $\frac{1}{4}\times 176=44$ g of protein.

Thus Shaun need to consume = (50-44) g = 6 of protein to reach the required amount.

Shaun must consume 6 more g of protein to reach the required amount.

5 0

3 years ago