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Taya2010 [7]
3 years ago
10

To what volume should 5.06 mL of an 7.83 M sulfuric acid solution be diluted in order to obtain a final solution that is 2.36 M?

Chemistry
1 answer:
Akimi4 [234]3 years ago
4 0

the answer is 16.8 mL H2SO4 (aq).

the formula for dilation is M1*V1=M2*V2, where M is the molarity and V is the volume.

so plugging in...

1. (7.83)(5.06)=2.36*V2

2. multiply 7.83 and 5.06 then divide that by 2.36

3. V2= 16.788050...

4. to three sigfigs (because the problem has a minimum of 3) is 16.8 mL H2SO4 (aq).

i hope this is correct! i did it without pencil and paper and i’m currently learning this too! if it’s wrong i’m so truly sorry :(

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An exhaled air bubble underwater at 290.
Vinil7 [7]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 290 kPa

Final pressure (P_{2}) = 104 kPa

Initial volume (V_{1}) = 18.9 ml

To find:

Final volume (V_{2})

We know;

From the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

   P × V = constant

   P ∝ \frac{1}{V}

From the above equation;

              \frac{P_{1} }{P_{2} }  = \frac{V_{2} }{V_{1} }

P_{1} represents the initial pressure of the gas

P_{2} represents the final pressure of the gas

V_{1} represents the initial volume of the gas

V_{2} represents the final volume of the gas

Substituting the values of the above equation;

                    \frac{290}{104} = \frac{V_{2} }{18.9}

             V_{2} = 52.7 ml

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

6 0
3 years ago
Suppose there are two known compounds containing the generic elements X and Y. You have a 1.00-g sample of each compound. One sa
Lapatulllka [165]

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason: 

If you look at the combining masses of X and Y in each of the two compounds, 

The first compound contains 0.25g of X combined with 0.75g of Y 
so the ratio (by mass) of X to Y = 1 : 3 

The second compound contains 0.33 g of X combined with 0.67 g of Y 
so the ratio (by mass) of X to Y = 1 : 2 

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!) 

The first compound will then contain 4g of X and 12g of Y 
The second compound will then contain 6g of X and 12g of Y 

<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6 
<span>or 2 : 3 </span>

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

The two compounds given with the plausible formula must therefore contain the same ratio.

8 0
3 years ago
How much energy is used to melt 44.33 g of solid oxygen?
Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

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