Heat_1: Get the ice to 0 degrees
Convert 7 kg to grams
7 kg [1000 grams / 1 kg] = 7000 grams
Heat needed to get the the ice from - 9 to 0
deltat = 0 - -9 = 9 degrees
m = 7000 grams
c = 2.1 joules/gram
Heat_1 = m*c*deltat
Heat_1 = 7000 * 2.1 * 9
Heat_1 = 132,300 joules
Heat_2: Melt the ice.
There is no temperature change. The formula is 333 j/gram
Formula: H = mass * constant
H = 7000 g * 333 J / gram
H = 2331000 joules
Heat_3: Total amount of Joules needed.
2331000 + 132300 = 2 463 300 joules
Convert to Megajoules
2 463 300 joules * 1 megajoule / 1000000 = 2.63 megajoules.
Lead Nitrate is highly soluble in water. 37.65g/100 mL at 20*C
Answer:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Explanation:
Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).
The equation can be written as follow:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
The above equation can be balance as illustrated below:
C₆H₁₂O₆ + O₂ —> CO₂ + H₂O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O
There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:
C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O
There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:
C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O
Now, the equation is balanced.
Answer:
C.) Argon
Explanation:
This is because ionisation energy increases as we move from left to right in a period. Argon presents in the right most column and Argon is a novel gas and has 8 electrons in outermost orbital. So, it is highly stable. I hope I helped! ^-^
Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
