How does the density of a gas affect the speed of sound

1 answer:

4 0

Increased density decreases the speed of sound in a medium. While increased density can mean increased rigidity, or stiffness, it is not always the case. Greater density can be due to each molecule or atom having more momentum, and being slower to respond to the vibration of its neighbor.

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Isn’t urgent the first 3 will be marked BRAINIEST

bearhunter [10]

Answer:

Al2(SO4)3 and Mg(OH)2

Explanation:

1. Al has a charge of 3-, and SO4 of 2-

when you cross multiply the charges you get

Al2 and (SO4)3

*the reason theres a bracket around the sulfate ion is that the charge 3 is not for oxygen only, but the entire sulphate ion*

Hence, Al2(SO4)3

2. Mg has a charge of 2- and OH of 1-

again cross multiply

Mg (you dont need to add the 1) and (OH)2

again, the bracket around OH means the charge appiles to Oxygen AND hydrogen

hence, Mg(OH)2

7 0

3 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:

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