A solution is prepared by dissolving 51 g of salt in 340 g of

The mass composition of a compound that assists in the coagulation of blood is 76.71% carbon, 7.02% hydrogen, and 16.27% nitroge

brilliants [131]

Answer : The empirical of the compound is, $C_{11}H_{12}N_2$

Explanation :

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 = 76.71 g

Mass of H = 7.02 g

Mass of N = 16.27 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{76.71g}{12g/mole}=6.39moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.02g}{1g/mole}=7.02moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.27g}{14g/mole}=1.16moles$

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

For C = $\frac{6.39}{1.16}=5.5$

For H = $\frac{7.02}{1.16}=6.0\approx 6$

For N = $\frac{1.16}{1.16}=1$

The ratio of C : H : N = 5.5 : 6 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : N = 11 : 12 : 2

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

The Empirical formula = $C_{11}H_{12}N_2$

Therefore, the empirical of the compound is, $C_{11}H_{12}N_2$

4 0

4 years ago

What is the atomic number of the yet-undiscovered element directly below francium ( Fr) in the periodic table?

kakasveta [241]

The atomic number of Francium is 87.

4 0

3 years ago

Please answerrrrrrrrrrrrrrrrrrrrrrrrr

nydimaria [60]

Answer: I'm in no way like an expert but I've heard water goes from an elevated area to a lower area. So it should be arrows that start at a more elevated to less elevated. Like white to brown, and orange to yellow, and stuff like that. I hope that helped? Anyway, have a lovely day!

Explanation:

8 0

3 years ago

A 2.5% (by mass) solution concentration signifies that there is of solute in every 100 g of solution. 2. therefore, when 2.5% is

Stels [109]

Answer #1. A 2.5% (by mass) solution concentration signifies that there is 2.5 grams of solute in every 100 g of solution.
To calculate 2.5% by mass solution, we divide the mass of the solute by the mass of the solution and then multiply by 100.

Answer #2. therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution.
This means that weighing out 2.5 grams of solute and then adding 97.5 grams of solvent would make a total of 100 gram solution:
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution

Answer#3. a solution mass of 1 kg is 10 times greater than 100 g, thus 1kg of a 2.5% ki solution would contain 25 grams of ki.
Since 1000 grams is 1 kg, we multiply 10 to each mass so that 100 grams becomes
1000grams:
mass of solute / mass of solution = 2.5g*10 / [(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution

6 0

4 years ago

b) 2C2H2(g) + 5O2(g)⟶4CO2(g) + 2H2O(l) 6.54 Calculate the heats of combustion for the following reactions from the standard enth

IRINA_888 [86]

Answer:

a. ΔH⁸ = -1420 kJ/mol b. ΔH⁸ = -1144.84 kJ/mol

Explanation:

a.

C₂H₄ (g) + 3 O₂ (g) ------------------------ 4 CO₂ (g) + 2 H₂O (l) ΔH⁸ = ?

ΔH⁸f kJmol 52.47 0 -399.5 -285.83

ΔH⁸ = 2(-399.5) + 2 (-285.83) - (52.47)

ΔH⁸ = -1420 kJ/mol

b.

H₂S (g) + 3 O₂ (g) ---------------------- 2 H₂O (l) + 2 SO₂ (g)

ΔH⁸f kJmol -20.50 0 -285.83 -296.84

ΔH⁸ = 2(-285.83) + 2 (-296.84) - (-20.50)

ΔH⁸ = -1144.84 kJ/mol

4 0

3 years ago