Molar mass of CH2NH2COOH - 75
Given mass of CH2NH2COOH - 30
Moles of CH2NH2COOH = Given mass/ Molar mass
moles of CH2NH2COOH = 30/75 = 0.4 mol
One mole of CH2NH2COOH contains 32 gram of oxygen
0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen
Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!
Answer:
The correct answer is 0.3582 M.
Explanation:
Based on the given information, the volume of KOH given is 19.55 ml, the volume of sulfuric acid given is 20.20 ml, and the molarity of sulfuric acid is 0.3467 M. There is a need to find the molarity of KOH.
The formula to use in the given case is,
M1V1 = M2V2
Here M1 is the molarity of KOH, V1 is the volume of KOH, M2 is the molarity of sulfuric acid, and V2 is the volume of H2SO4.
Here V1 is 19.55 ml, M2 is 0.3467 M, and V2 is 2020 ml.
Now putting the values in the equation we get,
M1 * 19.55 = 0.3467 * 20.20
M1 * 19.55 = 7.00334
M1 = 7.00334/19.55
M1 = 0.3582 M
Hence, the molarity of the given KOH is 0.3582.
Answer:
The mass of cryolite will be produced = 71247 g or, 71.247 kg
Explanation:
The balanced chemical equation for the synthesis of cryolite
Al₂O₃(s) + 6 NaOH(l) + 12 HF(g) → 2 Na₃AlF₆ + 9 H₂O(g)
Answer:
fundamental frequency in helium = 729.8 Hz
Explanation:
Fundamental frequency of an ope tube/pipe = v/2L
where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L is length of the pipe
To find the length of the pipe,
frequency = velocity of sound / 2L
272 = 340 / 2 L
L = 0.625 m
If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.
Also, the velocity of sound is inversely proportional to square root of molar mass of gas; v ∝ 1/√m
v₁/v₂ = √m₂/m₁
v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium
340 / v = √4 / 28.8
v₂ = 340 / 0. 3727
v₂ = 912.26 m /s
fundamental frequency in helium = v₂ / 2L
fundamental frequency in helium = 912.26 / (2 x 0.625)
fundamental frequency in helium = 729.8 Hz
Answer:the third one/ regulates the transfer of material into and out of the cell.
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