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3 years ago

1 Smurf = 3 Snorks 1 Snork = 6 Pookas 8 Pookas = 1 Flintstone1 Smurf = 3 Snorks

Chemistry

1 answer:

DIA [1.3K]3 years ago

5 0

Answer:

all together you have 13

Explantation:

you have to add all them together

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Water is boiled in a flask with a balloon over the top of the water gets the balloon expand what has happened to the weight of t

Verdich [7]

It decreases since a air less balloon is heavier than a Aired balloon

7 0

3 years ago

6. A sample of a gas has a mass of 0.527 g. Its volume is 0.35 L at a temperature of 88 degree Celsius and a pressure of 945 mm

Stels [109]

<u>Answer:</u> The molar mass of the gas is 35.87 g/mol.

<u>Explanation:</u>

To calculate the mass of gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of gas = 945 mmHg

V = Volume of the gas = 0.35 L

m = Mass of gas = 0.527 g

M = Molar mass of gas = ? g/mo

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of gas = $88^oC=[88+273]=361K$

Putting values in above equation, we get:

$945mmHg\times 0.35L=\frac{0.527g}{M}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 361K\\\\M=35.87g/mol$

Hence, the molar mass of the gas is 35.87 g/mol.

5 0

3 years ago

PLEASE HELP

Mars2501 [29]

Answer: it’s A

Explanation:

3 0

3 years ago

How many moles are in 25.0 grams of KMnO4?

Lilit [14]

Answer:- 0.158 moles

Solution:- Moles are calculated on dividing the grams by the molar mass.

25.0 grams of $KMnO_4$ are given and we are asked to calculate the moles.

For this we need the molar mass so let's calculate it first.

Molar mass is the sum of atomic masses of all the atoms present in the molecule.

Molar mass = $39.098+54.938+4(16)$

= 158.036 gram per mol

Now, we will divide the given grams by molar mass to get the moles and the set is shown as:

$25.0gKMnO_4(\frac{1mol}{158.036g})$

= $0.158molKMnO_4$

So, there are 0.158 moles of $KMnO_4$ .

3 0

3 years ago

Calculate the pH of a buffer solution prepared by mixing 200 mL of 0.10 M NaF and 100 mL of 0.050 M HF.

dlinn [17]

Answer:

$pH=3.74$

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one could compute the pH considering that the pKa of hydrofluoric acid, HF, is 3.14:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the concentration of the base and acid are computed by considering the mixing process with a total volume of 300 mL (0.3 L):

$n_{HF}=0.1L*0.05mol/L=0.005molHF\\\\n_{NaF}=0.2L*0.1mol/L=0.02molNaF$

$[HF]=\frac{0.005molHF}{0.3L}=0.017M$

$[NaF]=\frac{0.02molHF}{0.3L}=0.067M$

Therefore, the pH turns out:

$pH=3.14+log(\frac{0.067M}{0.017M} )\\\\pH=3.74$

Regards.

8 0

3 years ago