Answer: 1.997 M
Explanation:
molarity = moles of solute/liters of solution or
first we have to find our moles of solute (mol), which you can find by dividing the mass of solute by molar mass of solute
mass of solute: 92 g
molar mass of solute: 46.08 g/mol
let's plug it in:
next, we plug it into our original equation:
Answer:
Hydrogen
Explanation:
Hydrogen is the only non metal in group on of the period table of elements.
Answer:
Explanation:
Hello,
In this case, since the 3% by volume is represented as:
By using the ideal gas equation we compute the density of CO:
Then we apply the conversion factors as follows:
Regards.
<em />If 100 brownies require 6 eggs, 5 cups of flour, and 2 sticks of butter. Then, 50 brownies should require half of those required to make 100 brownies. Use ratio and proportion to determine the number of eggs needed:
100/50 = 6/x
x = 3<span />
Answer:
- <em>The average mass of calcium in each sample is: </em><u>0.978 g</u>
<em />
- <em>The absolute uncertainty is: </em><u>0.008 g</u>
Explanation:
The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.
When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.
The relative uncertainty is the absolute uncertainty divided by the quantity:
- Relative uncertainty = 0.1g / 12.2 g = 0.008
The average mass of calcium is calculated using proportions, along with the molar masses:
- Molar mass of calcium: 40.078 g/ mol (from a periodic table)
- Molar mass of calcite: 100.085 g/mol (given)
Proportion:
- 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite
- x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium
So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:
- Average mass = total mass of five samples / number of samples
- Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>
So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).
The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:
- Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g
The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.