Answer:
The efficiency rating of the jack is 0.067.
Explanation:
We have, a person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m.
It is required to find the the efficiency rating of the jack. It is equal to the ratio of output work to the ratio of input work. So,

Thus, the efficiency rating of the jack is 0.067.
Total weight of the mixture of sand and salt = 5.25 g
Weight of sand separated = 3.64 g
Weight of the salt in the mixture = (5.25 - 3.64) g
= 1.61 g
Then the percentage of sand in the sample = [(3.64/5.25) * 100] percent
= 69.33%
So the percentage of sand in the total mixture of sand and salt is 69.33%.
Answer:
1.10 × 10⁻¹² m
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
Speed of Light = Wavelength times Frequency
Explanation:
<u>Step 1: Define</u>
ν = 2.73 × 10²⁰ Hz
<u>Step 2: Find wavelength</u>
3.0 × 10⁸ m/s = λ(2.73 × 10²⁰ Hz)
λ = 1.0989 × 10⁻¹² m
<u>Step 3: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules.</em>
1.0989 × 10⁻¹² m ≈ 1.10 × 10⁻¹² m
Answer:
2.87 gram
N2 is the limiting agent
Explanation:
We will find out if there is sufficient N2 and h2 to produce NH3
a) For 2.36 grams of N2
Molar mass of N2 = 28.02
Number of moles of N2 in 2.36 grams = 2.36/28.02
Mass of NH3 = 17.034 g
Now NH3 produced form 2.36 grams of N2 =
2.36/28.02 * 2 * 17.034 = 2.87 g NH3
b) For 1.52 g of H2
NH3 produced = 1.52/2.016 * (2/3) * 17.034 = 8.56
N2 Is not enough to produce 2.87 g of NH3 and also H2 is not enough to make 8.56 g of NH3.
N2 is the limiting agent as it has smaller product mass
Answer:
29 mL
Explanation:
<h3>Equation</h3>
The question needs us to find the volume of the liquid. The equation for volume using density and mass is:
Volume = Mass / Density
<h3>Solve</h3>
We can substitute the given values for density and mass into the equation:


<h3>Additional Comments</h3>
The answer we obtained (29 mL) is rounded to two significant figures. When multiplying or dividing, the amount of significant figures in the final answer is always the least amount of significant figures in one of the values.
Below are the significant figure rules:
Nonzero digits will always be significant (eg. 54 --> 2 significant figures)
Zeroes at the beginning of a number will never be significant (eg. 0.1 --> 1 significant figure)
Zeroes between two nonzero digits will always be significant (eg. 504 --> 3 significant figures)
Zeroes following a number will always be significant if the number contains a decimal point (eg. 40.0 --> 3 significant figures)