Answer:
Divide the mass of the water lost by the mass of hydrate and multiply by 100.
Explanation:
i think
Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
Answer:
Answer is letter B
Explanation:
The first one is wrong because acids release H+, not bases.
The third one is wrong because the pH is exactly 7, not greater.
The last one is wrong because it is vague and does not fit a neutralization reaction.
If your options are among the following:
<span>0.64 M 1.0 M 0.32 M 0.16 M.
Then the correct answer is 0.64 M. I hope this is what you were looking for</span>
E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
a) M(N</span>₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
