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1 year ago

When rubbing alcohol and water are at room temperature, which one boils first

Chemistry

2 answers:

Ugo [173]1 year ago

6 0

Answer:

rubbing alcohol

mylen [45]1 year ago

5 0

Answer:

alcohol

Explanation:

Alcohol has a lower boiling point

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How many moles of Cl in 38.5 mol of CCl2F2?

Gennadij [26K]

Answer:

77 moles

Explanation:

The given chemical compound is CCl2F2 is called Dichlorodifluoromethane. In CCl2F2, we have:

one mole of a Carbon atom.

two moles of Chlorine atom; &

two moles of Fluorine atom.

Given that one mole of Chlorine = 38.5 moles.

Then, two moles will be:

= 2 × 38.5 moles

= 77 moles

5 0

2 years ago

How is the le chatelier's principle used to ensure the reaction will go to completion?

murzikaleks [220]

Answer:

This done such that when a reaction is in equilibrium and disturb by and external force which would prevent its completion its adjust so as to offset that external force and still go on to completion

4 0

2 years ago

What is the mass of arsenic that is contained in a 50.0 g sample if the concentration is 5 ppm?

faltersainse [42]

Answer: The mass of arsenic found in the sample is 0.25 mg

Explanation:

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

Concentration of arsenic = 5 ppm

Mass of sample = 50 g

Putting values in above equation, we get:

$5=\frac{\text{Mass of arsenic}}{50g}\times 10^6\\\\\text{Mass of arsenic}=\frac{5\times 50}{10^6}=0.00025g=0.25mg$

Conversion factor used: 1 g = 1000 mg

Hence, the mass of arsenic found in the sample is 0.25 mg

5 0

3 years ago

A 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?

oee [108]

Answer:

Mass of sample in mg = 15,285 mg

Explanation:

Given:

Volume of urine sample = 15 ml

Density of sample = 1.019 g/ml

FInd:

Mass of sample in mg

Computation:

Mass = density x volume

Mass of sample in mg = Volume of urine sample x Density of sample

Mass of sample in mg = 1.019 x 15

Mass of sample in mg = 15.285 gram

Mass of sample in mg = 15.285 x 1,000

Mass of sample in mg = 15,285 mg

8 0

2 years ago

Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9

EleoNora [17]

Answer : The current passing between the electrodes is, $1.056\times 10^{-2}A$

Explanation :

First we have to calculate the charge of sodium ion.

$q=ne$

where,

q = charge of sodium ion

n = number of sodium ion = $2.68\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

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

$q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C$

Now we have to calculate the charge of chlorine ion.

$q'=ne$

where,

q' = charge of chlorine ion

n = number of chlorine ion = $3.92\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

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

$q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C$

Now we have to calculate the current passing between the electrodes.

$I=\frac{q}{t}+\frac{q'}{t}$

$I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}$

$I=1.056\times 10^{-2}A$

Thus, the current passing between the electrodes is, $1.056\times 10^{-2}A$

4 0

2 years ago