The molarity of H3PO4 solution if 15.0 ml is completely neutralized by 38.5 ml of 0.15m naoh is calculated as follows
find moles of NaOH used = molarity x volume
= 38.5 x 0.15 = 5.775 moles
write the reacting equation
3NaOH + H3PO4 = Na3PO4 + 3H2O
from the equation the reacting ratio between NaOH to H3Po4 which is 3:1 the moles of H3PO4 is therefore = 5.775/3 = 1.925 moles
molarity of H3PO4 is therefore = moles /volume
= 1.925/15 = 0.128 M
Answer:
40.3∘C
Explanation:
At planet B;
Water boils = 180∘C
Water freezes = 50∘C
In this planet the temperature difference = 180 - 50 = 130 compared to earth where the temperature difference is; 100 - 0 = 100
This means;
130 ∘C = 100 ∘C
x ∘C = 31 ∘C
x = 31 * 130 / 100
x = 40.3∘C
Answer: Non polar solvents
Explanation:
Since with increasing the size of alkyl group hydrophobic nature increases and solubility in polar solvents decreases .
Hence Carboxylic acids with more than 10 carbon atoms, solubility is more in non polar solvents.
Answer:
Explanation:
The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction
<u>half-lives</u> <u> t/yr </u> <u>Remaining</u>
0 0 1
1 1.3 billion ½
2 2.6 ¼
3 3.9 ⅛
We see that after 2 half-lives, ¼ of the original mass remains.
Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.
Original mass = 4 × 2.10 g = 
Answer:
(a) I⁻ (charge 1-)
(b) Sr²⁺ (charge 2+)
(c) K⁺ (charge 1+)
(d) N³⁻ (charge 3-)
(e) S²⁻ (charge 2-)
(f) In³⁺ (charge 3+)
Explanation:
To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.
(a) |
I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).
(b) Sr
Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).
(c) K
K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).
(d) N
N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).
(e) S
S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).
(f) In
In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).