All categories

2 years ago

What is the total mass of the reactants in the equation below?

Chemistry

1 answer:

Keith_Richards [23]2 years ago

4 0

Answer:

266.7 g

Explanation:

Molar mass of AlCl3=133.3405 g

2AlCl3=2×133.34

=266.7 g

Please mark brainliest

You might be interested in

What is the density of a substance that weighs 340g and has a volume of 40cm3?

jeka94

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{340}{40} = \frac{17}{2} \\$

We have the final answer as

Hope this helps you

5 0

2 years ago

Why does a solid change to a liquid when heat is added

Nata [24]

Particles will have more energy and will vibrate really fast.

(Hope this helps)

4 0

2 years ago

Read 2 more answers

A physician has ordered 0.50 mg of atropine, intramuscularly. If atropine were available as 0.25 mg/mL of solution, how many mil

sladkih [1.3K]

If 0.25mg of atropine is in 1mL
so
0.50mg of atropine is in x

$x=\frac{0.5mg*1mL}{0.25mg}=2mL$

6 0

3 years ago

A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo

lara [203]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

$\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M$

6 0

2 years ago

How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0

2 years ago