All categories

3 years ago

I need help with this question. Will give 5 stars

Chemistry

2 answers:

Arisa [49]3 years ago

5 0

Answer:

yes

Explanation:

Harlamova29_29 [7]3 years ago

3 0

Answer:

What is the question?

Explanation:

You might be interested in

For a titration of 25 ml of naoh solution, 23.30 ml of hcl solution if used. what is the molarity of naoh

Eduardwww [97]

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

As you can see here, one mole of acid neutralizes one mole of base.

We use the concentration equation, which states that,

c=nv

n is the number of moles v is the volume of solution

Rearranging for moles, we get,

n=c⋅v

So, we have:

nNaOH=0.1 M⋅0.05 L

=0.005 mol

Since one mole of acid neutralizes one mole of base, then we must have: nHCl=nNaOH.

And so,

cHCl=nHClvHCl

=0.005 mol0.03 L

≈0.17 M

3 0

3 years ago

After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of

zzz [600]

Answer:

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.

Explanation:

The percentage of water in the sample is lower than expected.

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible:

If part of the sample is splashed from the crucible the mass of water detected will be less.

B. The dehydrated sample absorbed moisture after heating:

If the sample absorbs water after heating the percentage of water would be higher than expected.

C. The amount of the hydrate sample used was too small:

Depending on the sample size, different procedures can be chosen for analysis.

D. The crucible was not heated to constant mass before use:

In many occasions the crucible is heated next to the sample and not in previous form.

E. Excess heating caused the dehydration sample to decompose:

If the sample decomposes during heating, the analysis should be discarded.

success with your homework

4 0

3 years ago

If a hydrocarbon molecule contains a triple bond, its chemical name ends in

murzikaleks [220]

-yne. Hope this helps.

4 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago

4. How is the coordination number determined?

balandron [24]

A coordination number can be determined by the usage of an atom towards a molecule from seeing how many numbers of atoms would have to be combined together in an atom.

3 0

2 years ago