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AnnyKZ [126]
2 years ago
9

If I dilute 334.75 mL of 1.28 M lithium acetate solution to a volume of 822.18 ml, what

Chemistry
1 answer:
zlopas [31]2 years ago
3 0

Answer:

C₂ =  0.52 M

Explanation:

Given data:

Initial volume = 334.75 mL

Initial concentration = 1.28 M

Final volume = 822.18 mL

Final concentration = ?

Solution:

C₁V ₁     =     C₂V₂

By putting values,

1.28 M×334.75 mL = C₂×822.18 mL

C₂ =  1.28 M×334.75 mL / 822.18 mL

C₂ =  428.48 M.mL / 822.18 mL

C₂ =  0.52 M

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Which phrase describes a hypothesis? a. reviewing information related to the problem b. testing a factor, or variable, in an exp
monitta

Answer:

c. suggesting a solution to a problem after studying it carefully

Explanation:

Hypothesis generally refers to an assumption made. This assumption can either be true or false. It is a reasoning presented in order to explain a phenomenon and is the starting point for further investigation.

Among the options, the closest that best describes it is option C.

4 0
3 years ago
A gas mixture contains 10.0 mole% H2O (v) and 90.0 mole % N2. The gas temperature and absolute pressure at the start of each of
Rainbow [258]

Answer: (a). T = 38.2 °C     (b). V = 1.3392 cm³     (c). ii and iii  

Explanation:

this is quite easy to solve, i will give a step by step analysis to solving this problem.

(a). from the question we have that;

the Mole fraction of Nitrogen, yи₂ = 0.1

Also the Mole fraction of Water, yн₂o = 0.1

We know that the vapor pressure is equal to the partial pressure because the vapor tends to condense at due point.

ρн₂o = ṗн₂o

      = yн₂oP = 0.1 × 500 mmHg = 50 mmHg

from using Antoine equation, we apply the equation

logρн₂o = A - B/C+T

T = B/(A - logρн₂o) - C

  = 1730.63 / (8.07131 - log 50mmHg) - 223.426

T = 38.2 °C

We have that the temperature for the first drop of liquid form is 38.2 °C

(b). We have to calculate the total moles of gas mixture in a 30 litre flask;

   n  = PV/RT  

   n = [500(mmHG) × 30L] / [62.36(mmHGL/mol K) × 323.15K] = 0.744 mol

Moles of H₂O(v) is 0.1(0.744) = 0.0744 mol

Moles of N₂ is 0.9(0.744) = 0.6696 mol

we have that the moles of water condensed is 0.0744 mol i.e the water vapor  in the flask is condensed

Vн₂o = 0.0744 × 18 / 1 (g/cm³)

Vн₂o = 1.3392 cm³

Therefore, the  volume of the liquid water is 1.3392 cm³

(3). (ii) and (iii)

The absolute pressure of the gas and The partial pressure of water in the gas would change if the barometric pressure drops.

cheers i hope this helps!!!!

4 0
3 years ago
PLZ HELP PLZ SOMEONE PLZ How old do u need to work as a secret agent
zmey [24]

I THINK 18 OR 21 I AM SORRY IF I AM WRONG BRO

3 0
3 years ago
A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
Plss help me pass . I am so confused in chemistry !!
Lapatulllka [165]

Answer:

Explanation:

Ok so an atom is each ball. So in the first one there are 5 balls. In the second one there are 4 and so on. A molecule contains more than two balls. So they are all molecules. For the counting reactants and products, count how many balls are to the left of the arrow which is your number of reactants and count the balls to the right to find the number of product atoms.

3 0
3 years ago
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