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2 years ago

How many atoms of calcium, Ca would be equal to 5.472 X 104 moles Ca?

1 answer:

Montano1993 [528]2 years ago

5 0

Answer:
3.29 × 10²⁸ Atoms

Solution:
Moles and Number of atoms are related to each other as,

Moles = Number of Atoms / 6.022 × 10²³ ------- (1)

Data Given;
Moles = 5.472 ×10⁴

Number of Atoms = ?

Solving eq. 1 for Number of Atoms,

Number of Atoms = Moles × 6.022 × 10²³

Putting values,

Number of Atoms = 5.472 ×10⁴ mol × 6.022 × 10²³ atoms.mol⁻¹

Number of Atoms = 3.29 × 10²⁸ Atoms

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A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2and 0.973 M C2H4 and al

Musya8 [376]

Answer:

The value of $K_p$ is 0.02495.

Explanation:

Initial concentration of $SCL_2$ gas = 0.675 M

Initial concentration of $C_2H_4$ gas = 0.973 M

Equilibrium concentration of mustard gas = 0.35 M

$SCl_2 (g) + 2 C_2H_4 (g)\rightleftharpoons S(CH_2CH_2Cl)_2(g)$

initially

0.675 M 0.973 M 0

At equilibrium ;

(0.675-0.35) M (0.973-2 × 0.35) M 0.35 M

The equilibrium constant is given as :

$K_c=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}$

$=\frac{0.35 M}{(0.675-0.35) M\times ((0.973-2 × 0.35) M)^2}$

$K_c=14.45$

The relation between $K_p$ and $K_c$ are :

$K_p=K_c\times (RT)^{\Delta n}$

where,

$K_p$ = equilibrium constant at constant pressure = ?

$K_c$ = equilibrium concentration constant =14.45

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 20.0°C =20.0 +273.15 K=293.15 K

$\Delta n$ = change in the number of moles of gas = [(1) - (1 + 2)]=-2

Now put all the given values in the above relation, we get:

$K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}$

$K_p=6.2\times 10^{4}$

$K_p=0.02495$

The value of $K_p$ is 0.02495.

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2 years ago

What’s the difference between a physical and chemical property

irga5000 [103]

Answer:

One is hands on and one is watching

Explanation:

I just know my science

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2 years ago

What is the boiling point of a 1.5 m aqueous solution of fructose? the boiling point elevation constant of water is 0.515Â°c/m.

GREYUIT [131]

Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point elevation.
Kb - the ebullioscopic constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.

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2 years ago

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2 years ago

Estimate the ph of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate (na2hpo4) and 1.25 mole of hyd

GenaCL600 [577]

The HCl added = 1.25 moles

and the moles of Na2HPO4 = 1 mole

Now when acid is added in the given solution of Na2HPO4

One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4

Na2HPO4 + H+ ---> NaH2PO4

Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution

So this will result into formation of a buffer of phosphoric acid and NaH2PO4

NaH2PO4 + H+ ---> H3PO4

pKa of H3PO4 = 2.1

so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58

so the pH will be in between 2.1 to 7.2

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3 years ago