Hello!
At
Standard Pressure and Temperature, an ideal gas has a molar density of
0,04464 mol/L.So, we need to apply a simple conversion factor to calculate the density of Sulfur Dioxide using the molar mass of Sulfur Dioxide.
So, the Density of Sulfur Dioxide (SO₂) at STP is
2,8599 g/LHave a nice day!
Answer: A substance that produces hydroxide ions when placed in water is base.
Explanation:
Bases are the substance:
- Which gives negatively charged hydroxide(
) ions in aqueous solution.
- Which have pH value ranging from 7 to 14.
Where as acid gives positively charged hydronium ion(
) in aqueous solution.
Answer:
<em>17500 calories</em> of chocolate bars are needed to eat to gain 5 pounds.
Explanation:
We can use ratios to calculate the answer using the information given in the question.
1 pound : 3500 grams
5 pounds : x grams
As it is given that the individual is burning no calories, we do not have to factor in any additional numbers.
<u><em>Method</em><em> </em><em>A</em><em>:</em></u>
To go from 1 in the first ratio to 5 in the second ratio, they multipled 1 by 5. Hence, to go from 3500 in the first ratio to x in the second ratio, we must multiply by 5.
x = 3500 × 5
x = 17500
<em><u>Method B</u>:</em>
To solve for the answer x, we can convert the ratios into fractions.
1 / 5 = 3500 / x
3500 / x = 1 / 5
To make x the subject, multiply the denominator of the left fraction with the numerator of the right fraction and place it on the left side. Then multiply the numerator of the left fraction with the denominator of the right fraction and place it on the right side.
x = 5 × 3500
x = 17500
Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample
