All categories

3 years ago

Hi, my question is not answer, please

Chemistry

1 answer:

Oksanka [162]3 years ago

4 0

Okay, pls give brainliest because I answered the fastest and how are you?

You might be interested in

Scientific method worksheet

AfilCa [17]

number 9 is : Organized Data is a customizable information product versioning and management system used to track and store document, images and other products or information

5 0

4 years ago

Recrystallization Step: The procedure asks for using equal amount by mass of the crude product and the recrystallization solvent

zubka84 [21]

Answer : The volume (in mL) of 95% ethanol added would be, 1.39

Explanation :

As we are given that:

Mass of 95 % ethanol = 1.2 g

Density of 95 % ethanol = 0.86 g/mL

Formula used :

$\text{Volume of ethanol}=\frac{\text{Mass of ethanol}}{\text{Density of ethanol}}$

$\text{Volume of ethanol}=\frac{1.2g}{0.86g/mL}$

$\text{Volume of ethanol}=1.39mL$

Therefore, the volume (in mL) of 95% ethanol added would be, 1.39

4 0

3 years ago

Which of the substances below would dissociate into ions when dissolved in water? Select all that apply.

dusya [7]

Answer:

B. NaBr

D. KOH

Explanation:

Below is the solubility rules given for you knowledge.

Salts of

Group 1 elements are soluble( $Li^+, Na^+, K^+,Cs^+, Rb^+$ )
Ammonium ion is soluble ( $NH4^+$ )
The nitrate are generally soluble( $NO3^-)$

of Cl- , Br- , and I- are soluble, except Ag+ , Pb+2, and (Hg2)+2
most sulfate are soluble, except Ba+2, Ca+2,Pb+2, Ag+, Sr+2.
most hydroxide salts are only slightly soluble, except NH+4, Li+, Na+, K+
Most carbonates are insoluble (CO3 2-) Except group 1 and NH+4
most phosphate are insoluble except group 1 and NH+4

so using the rules above

NaBr , KOH are soluble, Pb(OH)2 is slightly soluble and AgCl is not soluble.

7 0

3 years ago

Mercury has a density of 13.6 grams/mL, what is this density in cg/L?

Firdavs [7]

It would be 4.6cgL not sure tho because I didint do that good in this

4 0

3 years ago

Help?

Ahat [919]

Answer:

$KClO_3$

Explanation:

Hello!

In this case, as we know the mass of the total sample, we can first compute the mass of oxygen:

$m_O=22.9g-7.33g-6.65g=8.92g$

Next, we compute the moles of each element:

$n_K=\frac{7.33g}{39.9g/mol}= 0.184mol\\\\n_{Cl}=\frac{6.65g}{35.45g/mol}=0.188mol \\\\n_O=\frac{8.92g}{16.00g/mol} =0.5575mol$

Now, we divide the moles by 0.184 moles, the fewest ones, to obtain:

$K=\frac{0.184}{0.184}=1.0 \\\\Cl=\frac{0.187}{0.184}=1.0\\\\O=\frac{0.5575}{0.184} =3.0$

Therefore, the empirical formula is:

$KClO_3$

Regards!

3 0

3 years ago