Answer:
Q = 802.6 J
Explanation:
Given data:
Specific heat capacity of water = 4.18 J/g.°C
Mass of water = 12 g
Initial temperature = 23°C
Final temperature = 39°C
Heat required = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 39°C - 23°C
ΔT = 16 °C
Q = 12 g× 4.18 J/g.°C × 16 °C
Q = 802.6 J
<h3>
Answer:</h3>
B Aqueous
<h3>
Explanation:</h3>
We are given the equation;
KBr (aq) + AgNO₃ (aq) → KNO₃ (aq) + AgBr (s)
- The equation shows a precipitation reaction or a double displacement reaction.
- Precipitation reaction because two soluble salts reacts to produce a precipitate as one of the product.
- Double displacement reaction because the salts exchange cations and anions to form new compounds
- In the equation;
- KBr is in aqueous state since all salts of potassium (K) are soluble in water.
- KNO₃ and AgNO₃ are also in aqueous state as all nitrates are soluble in water.
- AgBr on the other hand is in solid state since its a precipitate that is insoluble in water.
Molar mads of C2H5OH:
12*2+1*5+16+1
24+5+17
46
no of moles = mass in grams/molar mass
3.4=m/46
3.4*46=m
m=156.4
m=156g
Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
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