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2 years ago

Thank you for anyone that helps

Chemistry

2 answers:

Stolb23 [73]2 years ago

7 0

C) 6.75 x 10^-7

Hope it helped ya

ruslelena [56]2 years ago

6 0

The answer is C.

Hope this helps!

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HELP! What is the percent composition of carbon in caffeine (C₈H₁₀N₄O₂)? Show your work

Dominik [7]

Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol.

8 0

2 years ago

A 50.0 mL sample of 12.0 M HCl is diluted to 200 mL. What is true about the diluted solution?

kobusy [5.1K]

The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

V1 = 50mL
C1 = 12.0M
V2 = 200mL
C2 = ?

<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

$c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M$

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

Learn more on concentration of a solution here;

brainly.com/question/2201903

7 0

2 years ago

Ammonium phosphate ((NH4)3 PO4) is an important ingredient to many fertilizers. It can be made by reacting phosphoric acid (H3 P

lara31 [8.8K]

Answer:

15.35 g of (NH₄)₃PO₄

Explanation:

First we need to look at the chemical reaction:

3 NH₃ + H₃PO₄ → (NH₄)₃PO₄

Now we calculate the number of moles of ammonia (NH₃):

number of moles = mass / molecular wight

number of moles = 5.24 / 17 = 0.308 moles of NH₃

Now from the chemical reaction we devise the following reasoning:

if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄

then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄

X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄

mass = number of moles × molecular wight

mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄

4 0

2 years ago

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$