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3 years ago

In a sample of solid ba(no3)2 the ratio of barium ions to nitrate ions is

Chemistry

1 answer:

Nata [24]3 years ago

7 0

I believe the ratio is 1:2

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if you were to compare the mass of the products and reactants in a reaction, you would find that the mass of the products is alw

Salsk061 [2.6K]

If you were to compare the mass of the products and reactants in a reaction, you would find that the mass of the products is <span>equal to the mass of the reactants.</span>

3 0

3 years ago

2. What mass of water absorbs 6700 J of heat to raise the temperature from 283K to 318K?

Alexxandr [17]

Answer:

Q = mcT ...you can either substitute the molar heat capacity of water in the place of c or the specific heat capacity of water.

Explanation:

7 0

2 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:

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

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

3 years ago

nevsk [136]

Answer:

C4H90H

Explanation:

nskwjxnxnxnslq

6 0

3 years ago

Equation Given : Al^(3+) + Na3PO4 ==> 3Na^+ + AlPO4

Helga [31]

1 mols of Aluminium ion forms 1 mol aluminium phosphate

Molar mass of AlPO_4

27+31+16(4)
58+48
106u

Moles of AlPO_4

61µg/106
0.000061/106
5.75×10^{-7}
57.5µmol

Moles of Al3+=57.5µmol

3 0

2 years ago