How much heat (kJ) is absorbed from water in order for 5.50 moles NH4NO3 to dissolve in water?

Now molecules:

MArishka [77]

Answer:

a

Explanation:

6 0

3 years ago

If the caffeine concentration in a particular brand of soda is 1.87 mg/oz, drinking how many cans of soda would be lethal? Assum

Drupady [299]

2.77mg caffeine / 1oz12oz / 1canLethal dose: 10.0g caffeine = 10,000mg caffeine First, find how much caffeine is in one can of soda, then divide that amount by the lethal dose to find the number of cans. (2.77mg caffeine / 1oz) * (12oz / 1can) = 33.24mg caffeine / 1can. (10,000mg caffeine) * (1can / 33.24mg caffeine) = 300.84 cans. Since we can't buy parts of a can of soda, then we have to round up to 301 cans. Notice how all the values were set up as ratios and how the units cancelled.

4 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:

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