Answer:
A beam balance is an example of a first class lever.
Explanation:
A beam balance is an example of a first class lever. In a first class lever, the fulcrum is between the effort (force) and the load. The effort (force) moves over a large distance to move the load a smaller distance.
Other examples of first class lever are pliers, scissors, a crow bar, a claw hammer, a see-saw and a weighing balance etc.
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
Answer:
Buffers are solutions that resist changes in pH, upon addition of small amounts of acid or base. The can do this because they contain an acidic component, HA, to neutralize OH- ions, and a basic component, A-, to neutralize H+ ions. Since Ka is a constant, the [H+] will depend directly on the ratio of [HA]/[A-].
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For it to be the same element it must contain the same number of protons
The answer is: the mass of carbon is 420.6 grams.
m(C₈H₁₈) = 500 g; mass of octane.
M(C₈H₁₈) = 114.22 g/mol; molar mass of octane.
n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈).
n(C₈H₁₈) = 500 g ÷ 114.22 g/mol.
n(C₈H₁₈) = 4.38 mol; amount of octane.
In one molecule of octane, there are eight carbon atoms:
n(C) = 8 · n(C₈H₁₈).
n(C) = 8 · 4.38 mol.
n(C) = 35.02 mol; amount of carbon.
m(C) = 35.02 mol · 12.01 g/mol.
m(C) = 420.6 g; mass of carbone.
