Density of 2.7 g/ml and volume of 35.6 ml what is the mass

Chemistry

1 answer:

Anettt [7]3 years ago

3 0

The math is set up like

35.6 ml * 2.7 g/ 1 ml

which will leave you with

96.12 g

You might be interested in

An atom of which element has the greatest attraction for electrons in a chemical bond?

Katyanochek1 [597]

Answer is (4) - Se.

Among the given choices Se has the highest electronegativity value as 2.4 compared to others. Hence, Se shows <span>greatest attraction for electrons in a chemical bond.

</span>Electronegativity is a value that tells us how an atom can attract electrons towards itself. <span>If the electronegativity is high, then the attraction to the electrons is also high.
</span>

5 0

2 years ago

Read 2 more answers

Which of the following will have the slowest rate of diffusion at a given temperature?

Orlov [11]

So diffusion is inversely proportional to mass !
so as mass of the particle increases, diffusion decreases !

4 0

3 years ago

Can someone help me with this?

amm1812

Answer:

dnaq tester

Explanation:

5 0

3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$