$P_1=0.492\:atm\\V_1 = 45.8\:ml\\V_2 = ?\\P_1 = 0.954\:atm\\\\Boyle's Law = P_1V_1 = P_2V_2\\\\Lets \:make\: V \:subject\:of\:the\:formula\\\\\frac{P_1V_1}{P_2} = \frac{P_2V_2}{P_2} \\ \\V_2 =\frac{P_1V_1}{P_2} \\V_2 = \frac{0.492\times45.8}{0.954}\\ \\V_2 = \frac{22.5336}{0.954} \\\\V_2 = 23.620\\\\V_2 = 23.6$

4 0

3 years ago

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Which of the following would dissolve in water?<br> CCl4<br> LiCl<br> CH4<br> PCl6

babymother [125]

Like dissolves like
so water is polar

CCl4 is nonpolar
LiCl is polar
CH4 is nonpolar
PCl6 is nonpolar

so LiCl would dissolve

7 0

3 years ago

Read 2 more answers

Personal eyeglasses provide as much protection as Group of answer choices splash proof chemical goggles a face shield safety gla

MArishka [77]

Answer:

none of the above

Explanation:

6 0

3 years ago

Estimate the heat of neutralization of H3PO4 in kJ/mol given these data: Heat transferred = 2.5 x 102 J Moles of H3PO4 = 1.4 x 1

OlgaM077 [116]

Below are the steps to get the answers:

<span>1.) write out the balance equation 3NaOh+H3PO4->Na3PO4+3H2O

2.) You are given everything needed to calculate q=heat transfer=2.2*10^2, H3PO4 moles= 1.5*10^-3, NaOH moles=5.0*10^-3

3.) equation is deltaHneutraliztion=q/Moles of limiting reagent H3PO4 is limiting reagent because lowest moles, and is used up first

4.) Now plug in variables DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole Notice we had to convert J to kj, </span>

7 0

3 years ago