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3 years ago

Hi no Link please

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

5 0

Answer:

Tell him u like him and be respect his idea of what he say it’s a 50/50 chance he likes u. I really believe u can do it.

Explanation:

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Carlos is making phosphorus trichloride using the equation below. He uses 15.5 g of phosphorus and collects 50.9 g of phosphorus

Oxana [17]

Answer : The mass of chlorine reacted with the phosphorus is, 53.25 grams.

Explanation :

First we have to calculate the moles of phosphorus.

$\text{Moles of phosphorus}=\frac{\text{Mass of phosphorus}}{\text{Molar mass of phosphorus}}$

$\text{Moles of phosphorus}=\frac{15.5g}{31g/mol}=0.5mol$

Now we have to calculate the moles of $Cl_2$

The balanced chemical reaction is:

$2P+3Cl_2\rightarrow 2PCl_3$

From the balanced chemical reaction, we conclude that

As, 2 moles of phosphorous react with 3 moles of $Cl_2$

So, 0.5 moles of phosphorous react with $\frac{3}{2}\times 0.5=0.75$ moles of $Cl_2$

Now we have to calculate the mass of $Cl_2$

$\text{Mass of }Cl_2=\text{Moles of }Cl_2\times \text{Molar mass of }Cl_2$

Molar mass of $Cl_2$ = 71 g/mol

$\text{Mass of }Cl_2=0.75mol\times 71g/mol=53.25g$

Therefore, the mass of chlorine reacted with the phosphorus is, 53.25 grams.

4 0

4 years ago

Read 2 more answers

Is acetylene a homogeneous or heterogeneous mixture

kow [346]

Acetylene is a pure substance so it is homogeneous.

8 0

3 years ago

Large sharks eat many other marine life animals. The sharks and the animals they eat are all part of which level of organization

Law Incorporation [45]

Answer:

Community, ecosystem, population

Explanation:

6 0

3 years ago

Read 2 more answers

The complete combustion of octane, C8H18, the main component of gasoline, proceeds as follows: 2C8H18 (l) 25 O2(g) --> 16 CO2

Svet_ta [14]

The ratio of mole number of the reactants and products is equal to the coefficients. So the answer is a. 18.75 mol. b. 35.1 g. c. 1.38 * 10^5 g.

7 0

3 years ago

Read 2 more answers

You are asked to prepare 500 mL 0.200 M acetate buffer at pH 5.00 using only pure acetic acid ( MW=60.05 g/mol, pKa=4.76), 3.00

Vilka [71]

Answer:

You need to weight 6,005 g of acetic acid

Explanation:

Using Henderson-Hasselbalch formula you will obtain:

5,00 = 4,76 +log₁₀ $\frac{[Ac^-]}{[Acac]}$

Where Ac⁻ is the salt of acetic acid (Acac).

Solving:

1,738 = $\frac{[Ac^-]}{[Acac]}$ (1)

Also, yo know that:

0,200 M = [Ac⁻] + [Acac] (2)

Replacing (2) in (1):

[Acac] = 0,0730 M.

Thus:

[Ac⁻] = 0,127 M

The moles of each compound are:

Acac = 0,0730 M × 0,500 L = 0,0365 mol

Ac⁻ = 0,127 M × 0,500 L = 0,0635 mol

To prepare these moles it is necessary to use:

Acac + NaOH → AcNa + H₂O

The initial moles of Acac must be:

0,0365 moles + 0,0635 moles = 0,100 moles

To obtain 0,0635 moles of Ac⁻ you need to take this quantity of NaOH moles.

Thus, to obtain a acetate buffer of 5,00 you need to add 0,100 moles of acetic acid and 0,0635 moles of NaOH because This NaOH will react with acetic acid producing 0,0635 moles of Ac⁻ and surplus 0,0365 moles of acetic acid.

Now, to obtain 0,100 moles of acetic acid from pure acetic acid:

0,100 moles × $\frac{60,05 g}{1 mol}$ = 6,005 g

You need to weight 6,005 g of acetic acid

I hope it helps!

7 0

4 years ago