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Norma-Jean [14]
2 years ago
11

The density of silver metal bar is 10.5 g/cm3 and it occupies a volume space of 0.5l. What is its mass in kilograms (kg).

Chemistry
1 answer:
mojhsa [17]2 years ago
4 0

Density is equal to the ratio of mass to the volume.

The mathematical expression is given as:

Density = \frac{mass}{volume}

Density of silver metal bar=10.5 g/cm^{3}

Convert g/cm^{3} into g/L

1 cm^{3} = 0.001 L

Thus, density  = \frac{10.5}{10^{-3}} g/L

= 10.5\times 10^{3}g/L

Volume = 0.5 L

Put the values,

10.5\times 10^{3}g/L= \frac{mass}{0.5 L}

m= 10.5\times 10^{3}g/L\times 0.5 L

=5.25 \times 10^{3}g

Now, convert gram into kg

1 g = 0.001 kg

Therefore, mass in kg=  \frac{5.25 \times 10^{3}}{10^{-3}}

= 5.25 kg

Thus, mass of silver metal bar in kg=5.25 kg



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The final volume of the sample of gas V_{2} = 0.000151 m^{3}

Explanation:

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Final temperature T_{2} = 336 K

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Relation between P , V & T is given by

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2 years ago
A sample of a solid labeled as NaCl may be impure. a student analyzes the sample and determines that it contains 75 percent chlo
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A sealed vessel contains 0.200 mol of oxygen gas, 0.100 mol of nitrogen gas, and 0.200 mol of argon gas. The total pressure of t
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Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As
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Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

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[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

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