Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g
Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:
M / E = 4/3
If we cross multiply,
3M = 4E
Now, substituting the value of M from earlier and solving for E,
E = (3 * 0.3357) / 4
E = 0.2518
There are 0.2518 grams of hydrogen per gram of carbon in ethane.
Answer:
Following are the responses to the given choices:
Explanation:
- The RBC crenation is implied through NaCl by 2,67 percent(m/v) because that solution becomes hypertonic to RBC because of the water within the RBC that passes externally towards the outskirts. RBC thus shrinks.
- 1.13% (m/v), because the low concentration or osmotic that all this solution shows is hypotonic regarding RBC because of the water which has reached the resulting swelling in RBC.
- Distilled H2 implies hemolytic distillation.
- Glucose is indicated by crenation at 8.69 percent (m/v).
- 5.0% (m/v) glucose and 0.9% (m/v) (Crenation is indicated by NaCl.v)
Answer:
which one?
Explanation:
???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
Answer:
B
Explanation:
hope the picture helps you to understand:)
Answer:
1.26 × 10^-8 M
Explanation:
We are given;
Number of moles of mercury (i) chloride as 0.000126 μmol
Volume is 100 mL
We are required to calculate the concentration of the solution.
We need to know that;
Concentration is also known as molarity is given by;
Molarity = Number of moles ÷ Volume
Number of moles = 1.26 × 10^-10 Moles
Volume = 0.01 L
Therefore;
Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L
= 1.26 × 10^-8 M
Thus, the molarity of the solution is 1.26 × 10^-8 M