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2 years ago

What is a solute?

Chemistry

1 answer:

rusak2 [61]2 years ago

8 0

the answer is because when you research that would be the answer

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Which of these formulas is NOT consistent with the guidelines for creating ionic compounds? Mark all that apply.

iVinArrow [24]

Answer:

Explanation:

because boron and fluorine are both nonmetals and don't fit the guidlelines for creating ionic compounds

8 0

3 years ago

According to the law of conservation of energy, the energy released by the system must be transferred to and absorbed by the sur

Andrew [12]

True!
Energy released by system is absorbed by surroundings.

5 0

3 years ago

hich metal will displace nickel in a compound? A. Arsenic B. Tin C. Zinc D. Platinum

Bess [88]

Hello there,

The answer to your question is:

C. Zinc

Hope this answer helps you.

7 0

2 years ago

Hey!!! Chemistry Question about MOLES... Please help...

yulyashka [42]

Answer:

= 13.0 moles O2

Explanation:

1] Given the equation: 2C8H18 + 25 O2 ----> 16CO2 + 18H2O

a. How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide?

8.33 moles CO2 X

25mol O2

16mol CO2

= 13.0 moles O2

6 0

2 years ago

Read 2 more answers

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

2 years ago