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2 years ago

How many grams are in 2.06x10-4 moles of calcium phosphate

Chemistry

1 answer:

Orlov [11]2 years ago

3 0

Answer:

63.86 mg

Explanation:

Molar mass is the mass of 1 mol of compound / atom.

molar mass of the compound is - 310 g/mol

mass of 1 mol is - 310 g

therefore mass of 2.06 x 10⁻⁴ mol is - 310 g/mol x 2.06 x 10⁻⁴ mol = 0.06386 g

mass of calcium phosphate is - 63.86 mg

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PLS HELP Students in a chemistry class added 5 g of Zinc (Zn) to 50 g of hydrochloric acid (HCl). A chemical reaction occurred t

Vitek1552 [10]

Answer: In simplest case mass of reactants is same as mass of products.

Without thinking this question deeper, mass of ZnCl2 would be 49, but..

Explanation: Reaction should be Zn + 2 HCl ⇒ ZnCl2 + H2

Amount of zinc is 5 g / 65,38 g/mol = 0,076476 mol and amount

of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.

Althought HCl is needed 0.152 moles, zinc is an limiting reactant.

So it is possible to produce only 0.076476 mol Hydrogen and its mass

is 0.154 g. Mass of ZnCl2 would be 0.076476 mol · (65.38 + 2·35.45) =

10.42 g

4 0

2 years ago

If kc = 7.04 × 10-2 for the reaction: 2 hbr(g) ⇌h2(g) + br2(g), what is the value of kc for the reaction: 1/2 h2(g) + 1/2 br2 ⇌h

Kay [80]

At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2

at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769

7 0

3 years ago

Which of the following are likely to form an ionic bond

sattari [20]

Answer:

No question is posted.

But Ionic bond is mostly between metals live Na, K, Ca, Mg and strong non metals like Cl, O, F

Explanation:

3 0

2 years ago

Calculate mole fraction of benzene (70 g) having 30 gram of mass of carbon tetrahedral

Fynjy0 [20]

Moles of benzene

70/90=7/9=0.8mol

moles of CCl_4

30/154
0.2mol

Now

$\\ \rm\Rrightarrow \chi_{C_6H_6}$

$\\ \rm\Rrightarrow \dfrac{\chi_{C_6H_6}}{\chi_{sol^n}}$

$\\ \rm\Rrightarrow \dfrac{0.8}{0.8+0.2}$

$\\ \rm\Rrightarrow \dfrac{0.8}{1}$

$\\ \rm\Rrightarrow 0.8$

4 0

1 year ago

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$

Finally, the Gibbs free energy for the reaction at 298.15K is:

$\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol$

Best regards.

3 0

2 years ago