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Lady bird [3.3K]
3 years ago
8

A supercomputer, nicknamed Roadrunner, built by IBM for the Los Alamos National Labs can perform about 1.03 petaflop/s (1 petafl

op is 1015 calculations). Determine how many seconds it would take this computer to count a mole of things. Convert this figure into years.
Chemistry
1 answer:
Dmitrij [34]3 years ago
7 0

Answer:

18.5 years

Explanation:

(6 x 10^23)/(10^15 x 1.03 x 3600 x 24 x 365)

Hope this helps!

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The density of water at 30.0 °C is 0.9956 g/mL. If the specific gravity of acetic acid is 1.040 at 30.0 °C, what is the density
mash [69]

Answer:

The density of acetic acid at 30°C = 1.0354_g/mL

Explanation:

specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)

Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)

= 0.9956 g/mL × 1.040

= 1.0354_g/mL

Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.

Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure

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The movement of water both on and<br> above Earths surface
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4 0
3 years ago
7. If a waste container has 87.0 mL of 0.234 M acetic acid, how many milliliters of 2.00 M sodium
Naya [18.7K]

Answer:

V_{base}=10.2mL

Explanation:

Hello.

In this case, since the neutralization of the acid requires equal number of moles of both acid and base:

n_{acid}=n_{base}

Whereas we can express it in terms of concentrations and volumes:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we can compute the volume of sodium hydroxide (base) as follows:

V_{base}=\frac{M_{acid}V_{acid}}{M_{base}} \\\\V_{base}=\frac{87.0mL*0.234M}{2.00M}\\ \\V_{base}=10.2mL

Best regards.

8 0
3 years ago
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