Answer:
100 cg/1g
Step-by-step explanation:
1 cg = 0.01 g Multiply by 100
100 cg = 1 g
(a) is <em>wrong</em>. The correct conversion factor is 1000 cm³/1 L.
(b) is <em>wrong</em>. The correct conversion factor is 1000 mL/1 L.
(c) is <em>wrong</em>. The correct conversion factor is 1 m/10 dm.
Answer:
The average atomic mass of bromine is 79.9 amu.
Explanation:
Given data:
Percentage of Br⁷⁹ = 55%
Percentage of Br⁸¹ = 45%
Average atomic mass of bromine = ?
Formula:
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
Now we will put the values in formula.
Average atomic mass = [55 × 79] + [81 ×45] / 100
Average atomic mass = 4345 + 3645 / 100
Average atomic mass = 7990 / 100
Average atomic mass = 79.9 amu
The average atomic mass of bromine is 79.9 amu.
Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
<span>Three sources of error that might account for the differences
in the enthalpy of fusion include the room temperature how much’ long you stirred
and another thing that might make it have different results is how long the ice
was out for </span>
Answer:
The final temperature of water is 54.5 °C.
Explanation:
Given data:
Energy transferred = 65 Kj
Mass of water = 450 g
Initial temperature = T1 = 20 °C
Final temperature= T2 = ?
Solution:
First of all we will convert the heat in Kj to joule.
1 Kj = 1000 j
65× 1000 = 65000 j
specific heat of water is 4.186 J /g. °C
Formula:
q = m × c × ΔT
ΔT = T2 - T1
Now we will put the values in Formula.
65000 j = 450 g × 4.186 J /g. °C × (T2 - 20°C )
65000 j = 1883.7 j /°C × (T2 - 20°C )
65000 j/ 1883.7 j /°C = T2 - 20°C
34.51 °C = T2 - 20°C
34.51 °C + 20 °C = T2
T2 = 54.5 °C
