Answer:
The answer is "33.95 
".
Explanation:
Formula:
where
Given value:
convert temperature celsius (°C) to Kelvin (K):
After solving the value it will give:
= 33.95
Answer:
Δx ≥ 1.22 *10^-10m
Explanation:
<u>Step 1:</u> Data given
The E. coli bacterial cell has a mass of 1.80 fg ( = 1.80 * 10^-15 grams = 1.80 * 10^-18 kg)
Velocity of v = 8.00 μm/s (= 8.00 * 10^-6 m/s)
Uncertainty in the velocity = 3.00 %
E. coli bacterial cells are around 1 μm = 10^−6 m in length
<u>Step 2:</u> Calculate uncertainty in velocity
Δv = 0.03 * 8*10^-6 m/s =2.4 * 10^-7 m/s
<u>Step 3:</u> Calculate the uncertainty of the position of the bacterium
According to Heisenberg uncertainty principle,
Δx *Δp ≥ h/4π
Δx *mΔv ≥ h/4π
with Δx = TO BE DETERMINED
with m = 1.8 *10^-18 kg
with Δv = 2.4*10^-7
with h = constant of planck = 6.626 *10^-34
Δx ≥ 6.626*10^-34 / (4π*(1.8*10^-18)(2.4*10^-7))
Δx ≥ 1.22 *10^-10m
Answer:
a) 8.33 ml of the original stomach acid is neutralized
b) 191.67 ml of the stomach acid was neutralized
c) 249.68 ml acid would be neutralized by the original tablet
Explanation:
a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?
25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid
8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid
b) how much stomach acid was neutralized y the 4.3628 g tablet?
It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)
This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml
c) how much stomach acid would have been neutralized by the original 5.6832 g tablet
4.3628 g antacid is equivalent to 191.67 ml acid ( this is the answer for question 2)
5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid
Answer:
Liquids and gases flow easily because their particles can move or slide past one another.
Explanation:
Hope this helped Mark BRAINLEST!!
