Answer:
0.20 moles
Explanation:
The pressure is proportional to the quantity of gas at a given temperature and volume. So, the quantity needs to be increased by a factor of ...
(35 psi)/(29.2 psi) = 175/146 ≈ 1.19863
The fractional increase required is ...
1.19863 -1 = 0.19863
__
The quantity of air currently in the tire is ...
1 mol·519.67°R/(atm·23.6442 L) × (29.2/14.7 atm) × (11.6 L) / (45+459.67)°R
= 1.0035 mol
so we need to add ...
(fraction to add) × (current quantity) = amount to add
0.19863 × 1.0035 mol = 0.1993 mol = amount to add
About 0.20 moles of air must be added to the tire to bring the pressure up.
Answer:
Summer
Explanation:
Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.
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The concentration (in M) of hydroxide ions in a solution at 25°C with a pOH of 3.58 is 2.6 × 10⁻⁴ M.
pOH is the measure of basic nature of a solution by evaluating the [OH⁻] concentration.
It is the negative logarithm of the hydroxide ion concentration.
Given,
pOH = 3.58
Temperature = 25°C = 298K
At 25°C, the relation of pOH and [OH⁻] concentration is as follows:
∴ pOH = -log [OH⁻]
⇒ 3.58 = -log [OH⁻]
⇒log [OH⁻] = -3.58
⇒ [OH⁻] = antilog (-3.58)
⇒[OH⁻] = 2.6 × 10⁻⁴ M
The concentration (in m) of hydroxide ions in a solution at 25.0 °C with a pOH of 3. 58 is 2.6 × 10⁻⁴ M.
Learn more about pOH here, brainly.com/question/17144456
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Answer:
3.5 g
Explanation:
The density of water is 1 g/mL. The mass (m) corresponding to 20.0 mL is 20.0 g.
We can calculate the heat (Q) required to raise the temperature of 20.0 mL of water 1 °C (ΔT).
Q = c × m × ΔT = 1 cal/g.°C × 20.0 g × 1 °C = 20 cal
where,
c is the specific heat capacity of water
There are 160 calories in 28 g of Cheetos. The mass that releases 20 cal is:
20 cal × (28 g/160 cal) = 3.5 g
The balanced chemical reaction is written as :
Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:
120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
