There is 0.02538502095915 Moles in 5 grams of gold.
.energy mechanical needs it Answer:
Yes.
Explanation:
Answer:
0.73L
Explanation:
The following data were obtained from the question :
V1 = 0.65 L
P1 = 3.4 atm
T1 = 19°C = 19 + 273 = 292K
V2 =?
P2 = 3.2 atm
T2 = 36°C = 36 + 273 = 309K
The bubble's volume near the top can be obtain as follows:
P1V1 /T1 = P2V2 /T2
3.4 x 0.65/292 = 3.2 x V2 /309
Cross multiply to express in linear form as shown below:
292 x 3.2 x V2 = 3.4 x 0.65 x 309
Divide both side by 292 x 3.2
V2 = (3.4 x 0.65 x 309) /(292 x 3.2)
V2 = 0.73L
Therefore, the bubble's volume near the top is 0.73L
Answer:
There are 1000 mg in 1. g
There are 1000 g in 1 kg
Each students needs 2,250 mg of clay
Explanation:
In order to determine the amount of how many kilograms to order, the teacher will need to find out the total mass of clay required by the students and then the teacher will have to convert the total mass to the units required for the purchase.
The teacher would have to find out how many milligrams make one kilogram as follows
1 kg = 1000 g
1 g = 1000 mg
Therefore, 2,250 mg = 2250/1000 g = 2.25 g
2.25 g = 2.25/1000 kg = 0.00225 kg.
Answer:
B: Adding water, then adding solute
Explanation:
This is because, say you have a solution with a certain concentration.
If you add more water, it will become more diluted (less concentrated)
If you add more solute, it will become more concentrated.
Therefore if you add water and solute, it could cancel out, and the concentration would remain the same.
Hope this helps! Let me know if you have any questions/ would like anything further explained :)
