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3 years ago

I NEED HELP PLEASE, THANKS! :)

Chemistry

1 answer:

konstantin123 [22]3 years ago

7 0

Answer:

Gasoline isn't a pure substance.

Explanation:

A typical boiling range for gasoline is 100–400F. Gasoline has a vapor pressure of 7 psi in the summer, comparing it to water pressure it's less then 0.3 psi. For example if you spill water on a flat surface it might take a few minutes or hours to completly evapourate. Gasoline on the other hand it will evapurate within a few minutes, that's an example of volatility. Hope this helps

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A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask

leonid [27]

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

Answer: The concentration of barium acetate solution is 0.375 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Ba(CH_3COO)_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.125\times 1000}{350ml}$

$Molarity=0.357M$

Therefore, the concentration of solution is 0.375 mol/L

8 0

3 years ago

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat

Mashcka [7]

Answer:

0.26g of NaCl is the maximum mass that could be produced

Explanation:

Based on the reaction:

HCl + NaOH → NaCl + H₂O

Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl

To solve this question we need to find limiting reactant. The moles of limiting reactant = Moles of NaCl produced:

Moles HCl -Molar mass: 36.46g/mol-:

0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl

Moles NaOH -Molar mass: 40g/mol-:

0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH

As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.

The mass of NaCl is:

Mass NaCl -Molar mass: 58.44g/mol-:

0.0045 moles * (58.44g/mol) =

<h3>0.26g of NaCl is the maximum mass that could be produced</h3>

8 0

3 years ago

A mixture containing 20 mole % butane, 35 mole % pentane and rest

notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

$Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB$

Whereas $y$ accounts for the fractions at the outlet distillate and $x$ for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

$0.9=\frac{y_bD}{z_bF}$

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

$y_bD=0.9*z_bF=0.9*0.2*100mol=18mol$

The total distillate flow:

$y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol$

And the total bottoms flow:

$F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol$

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

$x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025$

Then, the molar fraction of pentane and hexane:

$x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422$

$x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553$

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0

3 years ago

Read 2 more answers

How do I find the molar mass of Copper (II) Nitrate?

QveST [7]

Answer:

The molar mass of copper (II) nitrate is 187.5 g/mol.

Explanation:

The molar mass is the mass of all the atoms in a molecule in grams per mole. To calculate the molar mass of a molecule, we first obtain the atomic weights from the individual elements in a periodic table. We then count the number of atoms and multiply it by the individual atomic masses.

8 0

3 years ago

Balancing oxidation-reduction reactions Mg+ N2—>Mg3N2

BartSMP [9]

Answer:

${ \sf{3Mg_{(s)} + N_{2(g)} →Mg _{3}N_{2(s)}}}$

3 0

3 years ago