Answer: A volume of 500 mL water is required to prepare 0.1 M 
from 100 ml of 0.5 M solution.
Explanation:
Given: 
= 0.1 M, 
= ?
= 0.5 M, 
= 100 mL
Formula used to calculate the volume of water is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M from 100 ml of 0.5 M solution.
Answer:
88,88 % de O y 11,11 % de H
Explanation:
La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:
<em>Masa molar agua:</em>
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Masa molar = 2 + 16 = 18g/mol
100g H2O * (1mol / 18g) = 5.556 moles H2O.
Moles de hidrógeno:
5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H
Moles Oxígeno = Moles H2O = 5.556 moles
La masa de hidrógeno es:
11.11mol * (1g/mol) 11.11g H
La masa de oxígeno es:
5.556 mol * (16g / 1mol) = 88.89g O
Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:
<h3>88,88 % de O y 11,11 % de H</h3>
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Answer:
New volume V2 = 92.7 Liter (Approx)
Explanation:
Given:
V1 = 106 l
T1 = 45 + 273.15 = 318.15 K
P1 = 740 mm
T2 = 20 + 273.15 = 293.15 K
P2 = 780 mm
Find:
New volume V2
Computation:
P1V1 / T1 = P2V2 / T2
(740)(106) / (318.15) = (780)(V2) / (293.15)
New volume V2 = 92.7 Liter (Approx)
P = m*v
p = momentum
m = mass
v = velocity
p = (37 kg + 18 kg)*(1.2 m/s)
p = (55 kg)*(1.2 m/s)
p = 66 kg-m / s
