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babymother [125]
3 years ago
10

If a liquid has a density of 1200 g/l what is it’s density in charge/ml

Chemistry
1 answer:
Juliette [100K]3 years ago
5 0

Answer:- 1.2\frac{g}{mL}

Solution:- The given density is 1200\frac{g}{L} and it asks to convert it to \frac{g}{mL} .

Here, we could notice that only volume unit is changing ad we know that, 1 L = 1000 mL. Let's make the set up as:

1200\frac{g}{L}(\frac{1L}{1000mL})

= 1.2\frac{g}{mL}

So, the density is 1.2\frac{g}{mL} .

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2. Using the following data, calculate the average atomic mass of magnesium (give your answer to the nearest
arlik [135]

Answer:

24.32

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 24

Abundance (A%) = 78.70%

Isotope B

Mass of B = 25

Abundance (B%) = 10.13%

Isotope C:

Mass of C = 26

Abundance (C%) = 11.17%

Average atomic mass of Mg =..?

The average atomic mass of Mg can be obtained as illustrated below:

Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

Average atomic mass = [(24 × 78.70)/100] + [(25 × 10.13)/100] + [(26 × 11.17)/100]

= 18.888 + 2.5325 + 2.9042

= 24.3247 ≈ 24.32

Therefore, the average atomic mass of magnesium (Mg) is 24.32

8 0
3 years ago
In which sample are the particles arranged in a regular geometric pattern?
andrew11 [14]
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7 0
3 years ago
How does the volume of water change the solubility of sodium chloride in simple words?
kupik [55]

Answer:

When sodium chloride dissolves in water to make a saturated solution there is a 2.5 per cent reduction in volume. ... The solubility of salt does not change much with temperature, so there is little profit in using hot water.

3 0
3 years ago
HELP PLEASE I HAVE A TEST TODAY AND I DON'T UNDERSTAND ANY OF THIS...
myrzilka [38]

Answer:

About 67 grams or 67.39 grams

Explanation:

First you would have to remember a few things:

 enthalpy to melt ice is called enthalpy of fusion.  this value is 6.02kJ/mol

  of ice  

 it takes 4.18 joules to raise 1 gram of liquid water 1 degree C

 water boils at 100 degrees C and water melts above 0 degrees C

 1 kilojoules is 1000 joules

  water's enthalpy of vaporization (steam) is 40.68 kJ/mol

  a mole of water is 18.02 grams

  we also have to assume the ice is at 0 degrees C

Step 1

Now start with your ice.  The enthalpy of fusion for ice is calculated with this formula:

q = n x ΔH    q= energy, n = moles of water, ΔH=enthalpy of fusion

Calculate how many moles of ice you have:

150g x (1 mol / 18.02 g) = 8.32 moles

Put that into the equation:

q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice

Step 2

To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C)  or 418 joules.

So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C.  418 x 150 = 62,700 joules or 62.7 kilojoules.

So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling.  That's a total of 112.79 kJ.

Step 3

The final step is to see how much energy is left to vaporize the water.

Subtract the energy you used so far from what you were told you have.

265 kJ - 112.79 kJ = 152.21 kJ

Again q = mol x ΔH (vaporization)

You know you only have 152.21 kJ left so find out how many moles that will vaporize.

152.21 kJ = mol x 40.68  or   mol = 152.21 / 40.68  = 3.74 moles

This tells you that you have vaporized 3.74 moles with the energy you have left.

Convert that back to grams.

3.74 mol   x  ( 18.02 g / 1 mol ) = 67.39 grams

5 0
2 years ago
El cation en el CaS​
Rus_ich [418]

Answer:

whatdidhesayyyy?

7 0
2 years ago
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