<span>We can use
the heat equation,
Q = mcΔT </span>
<span>Where Q is
the amount of energy transferred (J), m is the mass of the substance
(kg), c is
the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is
the temperature difference (°C).</span>
Let's assume that the finale temperature is T.
Q = 1200 J
<span>
m = 36 g
c = 4.186 J/g °C</span>
ΔT = (T -
22)
By applying
the formula,
1200 J = 36 g
x 4.186 J/g °C x (T - 22)
(T - 22) = 1200 J / (36 g x 4.186 J/g °C)
(T - 22) = 7.96 °C
T = (7.96 + 22) °C = 29.96 °C
T = 30 °C
Hence,
the final temperature is 30 °C.
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+).
Answer:
Explanation:
Mol of NaI = 0.405 mol
Molarity of solution = 0.724 M
Molarity is given by
The required volume is .
<span>Answer:
.01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.
so .01/.015 to .005/.015 ~ 67% D to 33% L.
And thus, the enantiomer excess will be 34%.</span>
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce 
g of (NH₄)₂S
25g of NH₃ will produce 
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
