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3 years ago

In a mixture of He, Ne, and Kr the total pressure is 125 kPa. The partial

Chemistry

1 answer:

tatiyna3 years ago

7 0

Answer:

Partial pressure of He = 73 kPa

Explanation:

Given:

Total pressure = 125 kPa

Partial pressure of Ne = 31 kPa

Partial pressure of Kr = 21 kPa

Find:

Partial pressure of He

Computation:

Total pressure = Partial pressure of Ne + Partial pressure of Kr + Partial pressure of He

125 kPa = 31 kPa + 21 kPa + Partial pressure of He

Partial pressure of He = 73 kPa

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Zolol [24]

Answer: the density is 997 kg

Explanation:

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3 years ago

what would the percentage of thorium-232 ( parent compound) be in a rock that was dated at 1.8 billion years

vovikov84 [41]

The half-life of Th-232 is 1.405 × 10¹⁰ years

Time elapsed = 2.8 x 10⁹ years

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A = A₀ = (1/2)^ t/t₁/₂

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3 years ago

Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the a

kiruha [24]

From the calculation, the molar mass of the solution is 141 g/mol.

<h3>What is the molar mass?</h3>

We know that;

ΔT = K m i

K = the freezing constant

m = molality of the solution

i = the Van't Hoft factor

The molality of the solution is obtained from;

m = ΔT/K i

m = 3.89/5.12 * 1

m = 0.76 m

Now;

0.76 = 26.7 /MM/0.250

0.76 = 26.7 /0.250MM

0.76 * 0.250MM = 26.7

MM= 26.7/0.76 * 0.250

MM = 141 g/mol

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5 0

2 years ago

What mass of oxalic acid dihydrate, H2C2O4•2H2O, is needed to make a 0.357 M solution of oxalic acid in a 250.0mL volumetric fla

GalinKa [24]

Molarity is expressed as the number of moles of solute per volume of the solution. The mass of oxalic acid dihydrate needed for the solution is calculated as follows:

Amount in moles: (0.357 mol H2C2O4•2H2O / L) (.250 L ) = 0.0893 mol H2C2O4•2H2O

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Hope this answers the question. Have a nice day.

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3 years ago

A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but additio

zhannawk [14.2K]

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate $SO_4~^-^2$ and the chloride $Cl^-$ :

<u>Sulfate</u>

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.( $Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2$ ), which are NOT soluble.

<u>Chloride</u>

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ( $Pb^+^2~,Ag^+~,Hg_2~^+^2$ ), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:

<u>"Calcium, strontium, or barium".</u>

I hope it helps!

7 0

3 years ago