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3 years ago

How many grams of CaCl2 are need to prepare 250 mL of a 0.15M solution?

Chemistry

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

Here is what I got from reading this question...

Explanation:

So if you dissolve 7.5625- g of CaCl2 in water, dissolve well and then add distilled in a 250- ml standard flask upto the mark and shake well for uniform Concentration, you will get 0.25- M CaCl2 solution.

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What type of reaction always takes one large molecule and breaks it down into its smaller pieces?

Ugo [173]

Answer:

The answer to your question is the letter D. a decomposition reaction

Explanation:

This is a brief description of the main chemical reactions.

a) A synthesis reaction is when two reactants are combined to form only one product.

b) A disynthesis reaction. I have not heard about this chemical reaction, I think it does not exist.

c) A combustion reaction is when an organic molecule reacts with oxygen to form carbon dioxide and water.

d) A decomposition reaction is when one reactant splits to form two or more products.

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3 years ago

Why are the chemicals in the water turning the phrogs gay?

Liula [17]

Answer:

Yes

Explanation:

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3 years ago

ATOMS WE LOVE ATOMS, answer this and get points , also thx for your help :)

LuckyWell [14K]

Answer:C because they have to I have the same mass before and after the equation.

Explanation:

HOPE THIS HELPED!! :) ;) <3<3

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2 years ago

Read 2 more answers

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

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3 years ago

It is common for students to overshoot the endpoint, meaning they add too much NaOH(aq) from the buret, which causes the solutio

umka2103 [35]

Answer: the percentage of acetic acid will be low.

Explanation: The major aim during titration of acids and bases is to determine the endpoint , that is exact point where the acid in the beaker changes colour, (in this case, pink )with an additional drop from the burette containing the base, since it is usually difficult to mark the equivalence point that tells us when all the substrate in the beaker has been neutralized completely with the buretted substance.

Overshooting the end point is an error which can occur when the person involved in the the titration accidently goes beyond this endpoint by adding too much of the substance(base) from the burette into the beaker missing the exact endpoint.

This implies that the person has added too much of the burreted liquid, ie the base than required , making the acid in the beaker to continue to react resulting to a lower concentration of the acid (acetic acid) with excess base.(NaOH)

8 0

3 years ago