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3 years ago

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Chemistry

1 answer:

crimeas [40]3 years ago

8 0

Answer:

Answer is A Peasants were valued members of society in Japan but not in Europe.

Explanation:

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PLEASE HELP WILL MARK BRAINLIEST!!!

lina2011 [118]

Answer:

#1. E

#2. A

#3. D

#4. F

#5. C

Please give me brainliest

5 0

3 years ago

Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H

lozanna [386]

Answer : The molecular formula of a caffeine is, $C_8H_{10}N_4O_2$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.12}{1.03}=4$

For H = $\frac{5.19}{1.03}=5.03\approx 5$

For N = $\frac{2.06}{1.03}=2$

For O = $\frac{1.03}{1.03}=1$

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_4H_5N_2O_1=C_4H_5N_2O$

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{194.19}{97}=2$

Molecular formula = $(C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2$

Therefore, the molecular of the caffeine is, $C_8H_{10}N_4O_2$

5 0

3 years ago

Corals absorb ______ from the ocean to form their skeletons.

Nina [5.8K]

<span>They absorb calcium ions and bicarbonate (hydrogen carbonate) ions.
Sooo B
I hope this helps! :)
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3 0

3 years ago

Read 2 more answers

When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by 10°C. What is the work done on

Molodets [167]

When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by 10°C. The work done on the gas is 845J.

The amount of work done on gas depends upon the internal energy change and the heat supplied to the system.

According to First Law of Thermodynamics, the change in internal energy is equal to the work done and the heat supplied to the system.

This is given by:

ΔU = W + Q

where, ΔU is change in Internal energy

W is the Work done

Q is the heat supplied

Given,

Q = 400J

Number of moles, n = 10

Change in temperature, ΔT = 10°C

Cv = 3/2R ; Since, the given gas is monoatomic (R=8.3)

We know that, ΔU = n Cv ΔT

On substituting the values in above formula,

ΔU = 10 × 3/2 × 8.3 × 10

ΔU = 1245J

Using,

ΔU = W + Q

1245J = W + 400

W = 845J

Hence, the work done on the gas is 845J.

Learn more about Thermodynamics here, brainly.com/question/1368306

#SPJ4

7 0

2 years ago

A certain ionic compound X has a solubility in water of 40.3 g/L at 20. degrees C. Calculate the mass X of required to prepare 5

tino4ka555 [31]

Answer:

20.1 g

Explanation:

The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.

So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:

40.3 g/L = m/V

40.3 g/L = m/0.5L

m = 40.3 g/L * 0.5L

m = 20.1 g

7 0

3 years ago