Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
Answer:
Cl₂O₇
Explanation:
For the reaction:
ClₓOₙ + H₂ → HCl + H₂O
Moles of HCl and moles of H₂O are:
HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl
H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O
As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:
6.39x10⁻³ mol Cl
2.236x10⁻² mol O
Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:
2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>
As empirical formula must be given in natural numbers, the empirical formula is:
<em>Cl₂O₇</em>
<em></em>
Answer:
The final temperature of water is 54.5 °C.
Explanation:
Given data:
Energy transferred = 65 Kj
Mass of water = 450 g
Initial temperature = T1 = 20 °C
Final temperature= T2 = ?
Solution:
First of all we will convert the heat in Kj to joule.
1 Kj = 1000 j
65× 1000 = 65000 j
specific heat of water is 4.186 J /g. °C
Formula:
q = m × c × ΔT
ΔT = T2 - T1
Now we will put the values in Formula.
65000 j = 450 g × 4.186 J /g. °C × (T2 - 20°C )
65000 j = 1883.7 j /°C × (T2 - 20°C )
65000 j/ 1883.7 j /°C = T2 - 20°C
34.51 °C = T2 - 20°C
34.51 °C + 20 °C = T2
T2 = 54.5 °C
Answer:
The answer to your question is: C₁₈ H₂₇ N O₃
Explanation:
Data
Carbon = 70.79 g
Hydrogen = 8.91 g
Nitrogen = 4.58 g
Oxygen = 15.72 g
Process
AT C = 12 g
AT H = 1 g
AT N = 14 g
AT O = 16 g
Carbon
12 g ------------------------ 1 mol
70.79 g ------------------------- x
x = (70.79 x 1) / 12
x = 5.9 mol of C
Hydrogen
1 g ----------------------- 1 mol
8.91 g --------------------- x
x = (8.91 x 1) / 1
x = 8.91 mol of H
Nitrogen
14 g ---------------------- 1 mol
4.58 g ------------------- x
x = (4.58 x 1) / 14
x = 0.33 mol
Oxygen
16 g ------------------------ 1 mol
15.72 g -------------------- x
x = (15.72 x 1)/16
x = 0.98
Divide by the lowest number of moles
Carbon 5.9 / 0.33 = 17.9 ≈ 18
Hydrogen 8.91 / 0.33 = 27
Nitrogen 0.33 / 0.33 = 1
Oxygen 0.98 / 0.33 = 2.9 ≈ 3
C₁₈ H₂₇ N O₃
I would he would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month. Hope it helps!!
