A liquid with a high viscosity cannot be a mixture.

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2

avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0

2 years ago

A pure gold bar is made of 19.55 mol of gold. What is the mass of the bar in grams

Law Incorporation [45]

<span>(19.55 mol Au) / ( 1 ) x (196.97 g Au) / ( 1 mol Au) =
19.55 x 196.97 =
3850.76 g Au

I hope this helps you and have a great day!! :)
</span>

4 0

1 year ago

Read 2 more answers

3. Sharon woke up on a sunny morning and ate breakfast. Then she looked outside and saw tall, quickly-forming clouds. The clouds

docker41 [41]

The change is that the air goes up then forms clouds.

7 0

2 years ago

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A metallic object holds a charge of −4.8 × 10−6 C. What total number of electrons does this represent? (e = 1.6 × 10−19 C is the

vagabundo [1.1K]

Answer:

$n=3.0\times 10^{13}$

Explanation:

Charge on 1 electron = $-1.6\times 10^{-19}\ C$

The expression for charge is:-

$Charge=n\times q_e$

Given that:- Charge = $-4.8\times 10^{-6}\ C$

$-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})$

$n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}$

Total number of electrons, n = $3.0\times 10^{13}$

5 0

1 year ago

If 42.1 mL of 1.02 M sodium hydroxide, measured using a graduated cylinder, is placed in a beaker filled with 300 mL of DI water

lys-0071 [83]

Answer:

Approximately 0.126 M

Explanation:

For the calculation of the dilution you take into account the moles of NaOH in the 42.1mL of the original solution and you use the new volume of 342.1 mL:

$(42.1 mL * 1.02 M ) Number Of Moles\\\\C=\frac{42.1mL*1.02M}{342.1 mL} = 0.126 M$

The standardization is necessary because a beaker is not not an instrument used to measure volumes and the marks on it only give an estimate of the volume of the solution, they are used to contain solutions and carry reactions among other things. If you would have measured the water with a graduated cylinder (an instrument designed to measure volumes) the standardization wouldnt be that necessary.

5 0

2 years ago