Answer: The steepness of a ramp affects it by making it easier or harder.
Explanation: It's a bit situational. If you were going up a steep ramp with a heavy load, it will increase the work necessary, whereas if you were going down a ramp, it would decrease the work necessary. If you need this simply put, think about biking up and down a hill. It would be easier going down than up.
Answer:
pH = 8.25
Explanation:
The acidity or basicity of a solution is measured by its pH. The pH scale ranges from 0 to 14. Solutions having pH from 0-6.9 are considered acidic, at 7 neutral and basic when ranging from 7.1-14.
pH is calculated as,
pH = - log [H⁺] ---- (1)
Where;
[H⁺] = concentration of Acid
Also, for bases pH i calculated using following formula,
pH = 14 - pOH
Therefore, Putting value of H⁺ in equation 1,
pH = - log [5.6 × 10⁻⁹]
pH = 8.25
The solution provided is basic in nature.
Answer:
2K +F₂→ 2KF
Explanation:
When we balance an equation, we are trying to ensure that the number of atoms of each element is the same on both sides of the arrow.
On the left side of the arrow, there is 1 K atom and 2 F atoms. On the right, there is 1 K and 1 F atom.
Since the number of K atoms is currently balanced, balance the number of F atoms.
K +F₂→ 2KF
Now, that the number of F atoms is balanced on both sides, check if the number of K atoms are balanced.
<u>Left</u>
K atoms: 1
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The number of K atoms is not balanced.
2K +F₂→ 2KF
<u>Left</u>
K atoms: 2
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The equation is now balanced.
Answer:
Oxygen's atomic weight is 16.00 amu. 1 mole of oxygen is 6.02 x 1023 atoms of oxygen 1 amu = 1.661 x 10-24g What is the molar mass (g/mole) of oxygen? Molar mass (in grams) is always equal to the atomic weight of the atom! Molar mass (in grams) is always equal to the atomic weight of the atom!
Answer:
P2 = 352 mm Hg (rounded to three significant figures)
Explanation:
PV = nRT
where P is the pressure,
V is the volume,
n is the moles of gas,
R is the gas constant,
and T is the temperature.
We must relate this equation to a sample of gas at two different volumes however. Looking at the equation, we can relate the change in volume by:
P1V1 = P2V2
where P1 is the initial pressure,
V1 is the initial volume,
P2 is the final pressure,
and V2 is the final volume.
Looking at this relationship, pressure and volume have an indirect relationship; when one goes up, the other goes down. In that case, we can use this equation to solve for the new pressure.
P1V1 = P2V2
(759 mm Hg)(1.04 L) = P2(2.24 L)
P2 = 352 mm Hg (rounded to three significant figures)
