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Zigmanuir [339]
3 years ago
12

Ann and Carol each lift the timber with a force of 80 kg/m/s2. If they lift the timber a distance of 1.25 meters, how much work

is each woman doing?
Question 1 options:

64.00 J


78.75 J


81.25 J


100.00 J
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
4 0
Work is defined energy transferred from one to another. 
The formula for work done is work done = force x distance

So in our problem, force is equal to 80 kg/ m / s^2 and distance is equal to 1.25 meters. So plugging in our values will give us:

work done = 80 kg/ m/ s^2 * 1.25 m
= 100.00 J is the answer.
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Answer:

a) 0.115 g

Explanation:

The balanced reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

To heat 55g of water, the energy in form of heat necessary can be calculated by:

Q = mcΔT

where Q is the heat, m is the mass, c is the specific heat (for water, c = 4.18 J/gºC), and ΔT is the variation of the temperature, which is 25ºC, so:

Q = 55x4.18x25

Q = 5747.5 J = 5.7475 kJ

So, for the reaction, 1 mol of CH₄ releases 802.3 kJ, so to release 5.7475 kJ will be necessary:

1 mol ---------------- 802.3 kJ

x ---------------- 5.7475 kJ

By a simple direct three rule:

802.3x = 5.7475

x = 7.164x10⁻³mol

The molar mass of CH₄ is : 12 (of C) + 4x1 (of H) = 16 g/mol

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6 0
4 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
ale4655 [162]

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

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