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2 years ago

5

What is the density of a solid with a mass of 75g and a volume of 5 cm^3?

1 answer:

beks73 [17]2 years ago

7 0

Answer: 15 g/cm^3

Explanation: formula for density is d= m/v where d is the density, m is the mass and v is the volume. If we plug in the values from the question, we get d= 75/ 5. 75 divided by 5 is 15 so the density is 15 grams per cubic centimetre.

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Use bond energies to determine Î´hrxn for the reaction between ethanol and hydrogen chloride. ch3ch2oh(g)+hcl(g)âch3ch2cl(g)+h2o

Alex

For this reaction to proceed, the following bond breaking should occur:
*one C-O bond* one H-Cl bond
After, the following bond formations should occur:*one C-Cl bond*one O-H bond
Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.
ΔHrxn = (1)(358) + (1)(431) + 1(-328) + 1(-463) =<em> -2 kJ</em>

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3 years ago

When electricity is produced from coal, the chemical energy in coal is first changed to thermal energy. True or False

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3 years ago

A hypothetical wave has 4.6 J of energy. What is its hypothetical, approximate frequency?

Mariulka [41]

Answer:

6.94 × 10^33Hz

Explanation:

E = hf

Where;

E = Energy of wave (J)

h = Planck's constant (6.626 × 10^-34J/s)

f = frequency (Hz)

According to the information provided in this question, the hypothetical energy of the wave is 4.6J

Hence, using E = hf

4.6 = 6.626 × 10^-34 × f

f = 4.6 / 6.626 × 10^-34

f = 0.694 × 10^34

f = 6.94 × 10^33Hz

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3 years ago

What is the SI base unit for time?

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2 years ago

A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume

pshichka [43]

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of $N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol$

amount of $SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol$

mole fraction of $N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330$

mole of fraction $SF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669$

PV = nRT

P of N_2 O $= \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa$

mole of SF_4 $=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa$

Total pressure = 39127.053 + 79126.36 = 118253.413 Pa

6 0

3 years ago