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luda_lava [24]
3 years ago
13

Make the conversion indicated ineach of the following:

Chemistry
1 answer:
Citrus2011 [14]3 years ago
4 0

Explanation:

(a) The length of a soccer field, 120 m (three significant figures), to feet.

1 m = 3.28084 feet

120 m=120\times 3.3 feet=396 feet

(b) The height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers

1 feet = 0.0003048 km

19,565 feet=19,565\times 0.0003048 km=5.963412 km

(c) The area of an 8.5 ×11-inche sheet of paper in cm^2.

Area of the sheet = 8.5 inches × 11 inches = 93.5 inches^2

1 inch = 2.54 cm

93.5 inches^{2}=93.5\times (2.54)^2 cm^2=603.2246 cm^2

(d) The displacement volume of an automobile engine, 161 inch^3, to liters.

1 inch^3 = 0.0163871 L

161 inch^3=161\times 0.0163871 L=2.63832 L

(e) The estimated mass of the atmosphere, 5.6\times 10^{15} tons, to kilograms.

1 ton = 907.185 kg

5.6\times 10^{15} tons=5.6\times 10^{15}\times 907.185 kg

=5.08023\times 10^{18} kg

(f) The mass of a bushel of rye,32.0 Ib, to kilograms

1 lb = 0.453592 kg

32.0 lb =32.0\times 0.453592 kg=14.515 kg

(g) The mass of a 5.00-grain aspirin tablet to milligrams.

(1 grain = 0.00229 oz)

Mass of 5 grains = 5 × 0.00229 oz = 0.01145 oz

1 oz = 28349.5 mg

0.01145 oz=0.01145\times 28349.5 mg=324.601775 mg

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Kepler's Law of Planetary Motion consists the movement of planets in the universe

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bagirrra123 [75]

a. pH=2.07

b. pH=3

c. pH=8

<h3>Further explanation</h3>

pH=-log [H⁺]

a) 0.1 M HF Ka = 7.2 x 10⁻⁴

HF= weak acid

\tt [H^+]=\sqrt{Ka.M}\\\\(H^+]=\sqrt{7.2.10^{-4}\times 0.1}\\\\(H^+]=8.5\times 10^{-3}\\\\pH=3-log~8.5=2.07

b) 1 x 10⁻³ M HNO₃

HNO₃ = strong acid

\tt pH=-log[1\times 10^{-3}]=3

c) 1 x 10⁻⁸ M HCl

\tt pH=-log[1\times 10^{-8}]=8

5 0
3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

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3 years ago
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ryzh [129]

Answer: The Answer is B. Energy can not be created nor destroyed.

Explanation:

4 0
3 years ago
Wastewater from a cement factory contains 0.280 g of Ca2+ ion and 0.0220 g of Mg2+ ion per 100.0 L of solution. The solution den
faltersainse [42]

<u>Answer:</u> The concentration of Ca^{2+}\text{ and }Mg^{2+} ions are 2.797 ppm and 0.212 ppm respectively.

<u>Explanation:</u>

To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Volume of gold = 100 L = 100000 mL    (Conversion factor:  1 L = 1000 mL)

Density of gold = 1.001 g/mL

Putting values in above equation, we get:

1.001g/mL=\frac{\text{Mass of solution}}{100000mL}\\\\\text{Mass of solution}=1.001\times 10^5g

To calculate the concentration in ppm (by mass), we use the equation:

ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6

  • <u>Calculating the concentration of calcium ions:</u>

Mass of Ca^{2+ ions = 0.280 g

Putting values in above equation, we get:

ppm(Ca^{2+})=\frac{0.280g}{1.001\times 10^5}\times 10^6=2.797ppm

  • <u>Calculating the concentration of magnesium ions:</u>

Mass of Mg^{2+ ions = 0.0220 g

Putting values in above equation, we get:

ppm(Mg^{2+})=\frac{0.0220g}{1.001\times 10^5}\times 10^6=0.212ppm

Hence, the concentration of Ca^{2+}\text{ and }Mg^{2+} ions are 2.797 ppm and 0.212 ppm respectively.

7 0
3 years ago
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