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arlik [135]
3 years ago
15

Mrs. Sail's class wanted to observe a local pond. The class collected some pond water to analyze. When they placed a drop of the

water under a microscope, they saw many microorganisms in the water. The pond water looked clear when they collected it in the jar, so what could explain the presence of these organisms? A) They must have grown in the jar overnight. B) The microorganisms are too small to be seen with the naked eye. C) Somehow, the jar was contaminated on the way from the pond to the class. D) The microorganisms only inhabit a very small section of the lake, and the class got lucky.
Chemistry
2 answers:
olasank [31]3 years ago
8 0

B) the microorganisms are to small too see with the naked eye

LiRa [457]3 years ago
3 0

B. its more of a reliable answer because it's true. You cant see it with the naked eye if you had to put in under a microscope.



I might be wrong.

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What is the molarity of solution that is 5.50 percentage by mass oxalic acid and has a density of 1.024 g/ml
Y_Kistochka [10]

Answer:

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.

Explanation:

Mass percentage of oxalic acid = 5.50%

This means that in 100 grams of solution there are 5.50 grams of oxalic acid.

Mass of solution , m = 100

Volume of the solution = V

Density of the solution = d = 1.024 g/mL

V=\frac{m}{d}=\frac{100 g}{1.024 g/mL}=97.66mL

V = 97.66 mL = 0.09766 L

(1 mL = 0.001 L)

Moles of oxalic acid = \frac{5.50 g}{90 g/mol}=0.06111 mol

Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}

The molarity of the solution :

=\frac{0.06111 mol}{0.09766  L}=0.6257M

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.

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3 years ago
9. In a titration experiment 34.7 of a 0.145M solution of barium hydroxide \ Ba(OH) 2 \ is added to 20mL of hydrochloric acid (H
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Answer:

Titration reveals that 12 mL of 1.5 M hydrochloric acid are required to neutralize 25 mL calcium hydroxide solution . What is the molarity of the Ca(OH) 2 solution

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2 years ago
If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at 127oC in an evacuated
zhuklara [117]

Answer : The the partial pressure of the nitrogen gas is 0.981 atm.

The total pressure in the tank is 2.94 atm.

Explanation :

The balanced chemical reaction will be:

(CH_3)_2N_2H_2(l)+2N_4O_4(l)\rightarrow 3N_2(g)+4H_2O(g)+2CO_2(g)

First we have to calculate the moles of dimethylhydrazine.

Mass of dimethylhydrazine = 150 g

Molar mass of dimethylhydrazine =60.104 g/mole

\text{Moles of dimethylhydrazine}=\frac{\text{Mass of dimethylhydrazine}}{\text{Molar mass of dimethylhydrazine}}

\text{Moles of dimethylhydrazine}=\frac{150g}{60.104g/mole}=2.49mole

Now we have to calculate the moles of N_2 gas.

From the balanced chemical reaction we conclude that,

As, 1 mole of (CH_3)_2N_2H_2 react to give 3 moles of N_2 gas

So, 2.49 mole of (CH_3)_2N_2H_2 react to give 2.49\times 3=7.47 moles of N_2 gas

Now we have to calculate the partial pressure of nitrogen gas.

Using ideal gas equation :

PV=nRT\\\\P_{N_2}=\frac{nRT}{V}

where,

P = Pressure of N_2 gas = ?

V = Volume of N_2 gas = 250 L

n = number of moles  N_2 gas = 7.47 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 127^oC=273+127=400K

Putting values in above equation, we get:

P_{N_2}=\frac{(7.47mole)\times (0.0821L.atm/mol.K)\times 400K}{250L}=0.981atm

Thus, the partial pressure of the nitrogen gas is 0.981 atm.

Now we have to calculate the total pressure in the tank.

Formula used :

P_{N_2}=X_{N_2}\times P_T

P_T=\frac{1}{X_{N_2}}\times P_{N_2}

P_T=\frac{1}{(\frac{n_{N_2}}{n_T})}\times P_{N_2}

P_T=\frac{n_{T}}{n_{N_2}}\times P_{N_2}

where,

P_T = total pressure = ?

P_{N_2} = partial pressure of nitrogen gas = 0.981 atm

n_{N_2} = moles of nitrogen gas = 3 mole  (from the reaction)

n_{T} = total moles of gas = (3+4+2) = 9 mole  (from the reaction)

Now put all the given values in the above formula, we get:

P_T=\frac{9mole}{3mole}\times 0.981atm=2.94atm

Thus, the total pressure in the tank is 2.94 atm.

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