You can automatically rule out CH₄ since it has no lone pairs at all around the central atom. Water has 2. Ammonia is the only Lewis structure that contains one lone pair.
Answer:
The answer to your question is: 6.55 x 10 ²³ atoms of Br
Explanation:
CH2Br2 = 37.9 g
MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g
174 g of CH2Br2 ------------------ 160 g of Br2
37.9 g of CH2Br2 --------------- x
x = 37.9 x 160/174 = 34.85 g of Br
1 mol of Br ----------------- 160 g Br2
x ---------------- 174 g Be2
x = 174 x 1 /160 = 1.088 mol of Br2
1 mol of Br ----------------- 6.023 x 10 ²³ atoms
1.088 mol of Br ------------- x
x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms
Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Explanation:
1) Data:
Water ⇒ C = 1 cal/g°C
m = 65.8 g
Ti = 31.5°C
Tf = 36.9°C
Heat, Q = ?
2) Formula:
Q = mCΔT
3) Calculations:
Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal
4) You can convert from calories to Joules using the conversion factor:
1 cal = 4.18 J
⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
Answer:
<h2>Hi there !</h2>
<h2>C. HCl</h2>
Explanation:
<h2>Reason :-</h2>
<h2>Salts are strong electrolytes, so they undergo complete dissociation.</h2><h3>Hope it helps u.....</h3><h3>Stay safe, stay healthy and blessed</h3><h3>Have a good day</h3><h3>Thank you ~</h3>
