The answer is 0.0171468704904. We assume you are converting between moles Mg(OH)2 and gram. This compound is also known as Magnesium Hydroxide. 1 mole<span> is equal to </span>1 moles<span> Mg(OH)2, or 58.31968 grams.</span>
Answer:
We need 8.7 mL of the stock solution
Explanation:
Step 1: Data given
Volume of the solution he wants to prepare = 50.0 mL = 0.050 L
Concentration of the solution he wants to prepare = 0.80 M
The concentration of the stock solution = 4.6 M
Step 2: Calculate the volume of the stock solution
C1*V1 = C2*V2
⇒with C1 = the concentration of the stock solution = 4.6 M
⇒with V1 = the volume of the stock solution = TO BE DETERMINED
⇒with C2 = the concentration of the prepared solution = 0.80 M
⇒with V2 = the volume of the prepared solution = 0.050 L
4.6 M * V2 = 0.80 M * 0.050 L
V2 = (0.80 M * 0.050 L) / 4.6M
V2 = 0.0087 L = 8.7 mL
We need 8.7 mL of the stock solution
Answer:
B
Explanation:
when present
double bond the copound is alkene
Answer:
We colect 11.2 L of gas
Explanation:
Let's apply the Ideal Gases Law to solve this:
P . V = n . R .T
where P and T in STP are 1 atm and 273°K
The thing is n which means the number of moles for the gas.
As we know, 1 mol of anything has 6.02x10²³ particles so
6.02x10²³ are occupied in 1 mol of gas
3.00x10²³ are occupied in (3.00x10² .1) / NA = 0.500 moles
So let's go to the formula:
1 atm . V = 0.500m . 0.082 . 273K
V = (0.500m . 0.082 . 273K) / 1atm
V = 11.2L
There is a rule, which says that 1 mol of gas in STP, occupies 22.4L so, since we have half a mole, it will occupy half the volume
Answer:
Explanation:
You have an acid that is acidic or a base that is basic. When you mix the two, they form water (assuming those are bronsted-lowry acids and bases) which is neutral.
