Answer:
thats one hint. but for 1 it will produce lava
Explanation:
number 2
some of the Earth's grandest mountains are composite volcanoes--sometimes called stratovolcanoes. They are typically steep-sided, symmetrical cones of large dimension built of alternating layers of lava flows, volcanic ash, cinders, blocks, and bombs and may rise as much as 8,000 feet above their bases
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
it's subduction
Explanation:
i know this because I just do lol
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
Answer:
Q = 2647 J
Explanation:
Specific heat capacity is the amount of energy required by one Kg of a substance to raise its temperature by 1 °C.
In thermodynamics the equation used is as follow,
Q = m Cp ΔT
Where;
Q = Heat = ?
m = mass = 660 g
Cp = Specific Heat Capacity = 0.3850 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = 23.35 °C - 12.93 °C = 10.42 °C
Putting values in eq. 1,
Q = 660 g × 0.3850 J.g⁻¹.°C⁻¹ × 10.42 °C
Q = 2647 J
