Answer:
19
Explanation:
Once you add all the weights together, you end up with 114 then you divide it by how many dogs there are, 6. 114/6 = 19. The average is 19 lbs.
Unit of M is also mole/L, where mole is the moles of solute and L is the volume of the solution. The latter is given: 158 mL or 0.158 L. So we need to find out the moles of NH4Br.
Moles of NH4Br = Mass of NH4Br/molar mass of NH4Br = 17.0g/(14+1*4+79.9)g/mol = 0.1736 mole.
So, the molarity of the solution = 0.1736mole/0.158L = 1.10 mole/L = 1.10 M
Answer:
30.17 × 10²³ atoms
Explanation:
Given data:
Number of moles of lead = 5.01 mol
Number of atoms = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
In given question:
1 mole = 6.022 × 10²³ atoms
5.01 mol × 6.022 × 10²³ atoms / 1 mol
30.17 × 10²³ atoms
Answer:
V equals four-thirds times pi times r cubed
Explanation:
Volume = a³ , where a is length of each side. Volume = l × w × h , where l is length, w is width and h is height. Volume = 4/3 πr³ , where r is the radius. Volume = πr²h , where r is the radius and h is the height.
Answer:
32.8%
Explanation:
All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).
First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:
- 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄
There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.
We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:
- 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb
Finally we calculate the percentage composition of Pb:
- 0.083 g Pb / 0.254 g salt * 100% = 32.8%
