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3 years ago

How many moles of xenon gas are necessary to exert a 0.64atm pressure in an 8.5L container at 300.0K?

Chemistry

2 answers:

navik [9.2K]3 years ago

5 0

0.026 moles is the right answer.

Sindrei [870]3 years ago

3 0

Answer:

0.026 moles

Explanation:

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Please only answer if you know the actual thing, don't search it up! Are these correct? :P

makkiz [27]

Hey buddy I am here to help!

1. C

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4 0

3 years ago

Read 2 more answers

A solution is made by adding 29.1 mL of concentrated perchloric acid ( 70.5 wt% , density 1.67 g/mL ) to some water in a volumet

Lera25 [3.4K]

Answer:

The concentration of the solution is 1.364 molar.

Explanation:

Volume of perchloric acid = 29.1 mL

Mass of the solution = m

Density of the solution = 1.67 g/mL

$m=1.67 g/mL\times 29.1 mL=48.597 g$

Percentage of perchloric acid in 48.597 solution :70.5 %

Mass of perchloric acid in 48.597 solution :

= $\frac{70.5}{100}\times 48.597 = 34.261 g$

Moles of perchloric acid = $\frac{34.261 g}{100.46 g/mol}=0.3410 mol$

In 29.1 mL of solution water is added and volume was changed to 250 mL.

So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)

$Molarity=\frac{Moles}{Volume (L)}$

$=\frac{0.3410 mol}{0.250 L}=1.364 M$

The concentration of the solution is 1.364 molar.

6 0

3 years ago

Use your calculator to determine the answer to the following calculation:<br> 1.0×10^−15/4.2×10^−7

lianna [129]

1.0×10^−15/4.2×10^−7=<span>2.3809524e-23 Hoped I helped!</span>

8 0

3 years ago

Given that 7.25 moles of carbon monoxide gas are present in a container of volume 11.90 L, what is the pressure of the gas (in a

d1i1m1o1n [39]

Answer:17.955atm

Explanation:Pv=nrt

P= nrt/v

P= 7.25*0.08205*360/11.90

P= 214.1505/11.90

P=17.995atm

6 0

3 years ago

How many sulfur atoms are present in 25.6 g of Al2(S2O3)3

IgorC [24]

Given the mass of $Al_{2}(S_{2}O_{3})_{3}$ =25.6 g

The molar mass of $Al_{2}(S_{2}O_{3})_{3}$ =390.35g/mol

Converting mass of $Al_{2}(S_{2}O_{3})_{3}$ to moles:

$25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}$

Converting mol $Al_{2}(S_{2}O_{3})_{3}$ to mol S:

$0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS$

Converting mol S to atoms of S using Avogadro's number:

1 mol = $6.022*10^{23}atoms$

$0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms$

5 0

3 years ago