<em>Answer:</em>
- 0.052301 km have 5 significant figure
- 400 cm have 1 significant figure
- 50.0 m have 3 significant figure
- 4500.01 ml have 6 significant figure
<em>Explanation:</em>
According to rules of significant figure
0.052301 km have 5 significant figure:
- Zero to the left of the first non zero digit not significant.
- Zero between the non zero digits are significant.
<em>400 cm have 1 significant figure:</em>
- Trailing zeros are not significant in numbers without decimal points.
<em>50.0 m have 3 significant figure:</em>
- Trailing zeros are significant in numbers when there is decimal points.
<em>4500.01 ml have 6 significant figure:</em>
- Zero between the non zero digits are significant.
Answer:
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
Answer: 9.04 g of H2O
Explanation:
First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O
The average atomic mass of oxygen is the atomic mass with respect to 1/12th the mass of a C-12 atom, related to natural abundance.
amu - atomic mass units
O-16 - 15.995 amu
O-17 - 16.999 amu
O-18 - 17.999 amu
the average atomic mass is 15.9994 amu
average atomic mass = ∑atomic mass of each isotope * relative abundance percentage
in other words the average atomic mass is closer to the atomic mass of the most abundant isotope. In this case 15.9994 is closest to 15.995 amu which is the atomic mass of O-16.
This means that O-16 is the most abundant isotope.
<span>I'd go with D. as it sounds fair and most common way to find out who would be best for the authority
</span><span>d)hold a free and fair election </span>
Answer:
c or d I think but I'm not completely shure
