Heat required = Q = 40 kcal
<h3>Further explanation</h3>
Given
mass of 500 g ice
Required
Heat required
Solution
The heat to change the phase can be formulated :
- Q = m.Lf (melting/freezing)
- Q = m.Lv (vaporization/condensation)
Lf=latent heat of fusion
Lv=latent heat of vaporization
Lf for water = 334 kj/kg=6.01 kJ/mol = 80 cal/g
Phase change(ice to water)
Q= 500 g x 80 cal/g
Q = 40 kcal
Answer:
B) H2SO4 (aq) + Ca(OH)2 (aq) → CaSO4 (aq) + 2 H2O(l)
Explanation:
A is a reaction between a salt FeCl3 and a base KOH
C is a n acid decomposing on it's own to form two products
D is mercury, a metal reacting with oxygen. Two elements reacting. Neither are an acid or a base
E. is an acid reacting with a metal to liberate hydrogen. There is no base
Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>
<em />
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
<em>Moles HCl:</em>
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
<em>Where P is pressure = 1atm assuming STP</em>
<em>V volume in L</em>
<em>n moles = 1.25x10⁻³ moles CO₂</em>
<em>R gas constant = 0.082atmL/molK</em>
<em>T = 273.15K at STP</em>
<em />
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
<h3>A. 30cm³</h3>
Answer:
About 547 grams.
Explanation:
We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.
To do so, we can use the initial value and convert it to grams using the molar mass.
Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:
Dimensional Analysis:
In conclusion, about 547 grams of copper (II) bicarbonate is produced.
Answer:
Q9. The independent variable in this experiment is the fertilizer. It is independent because she manipulating the variable to compare the growth.
Q10. The dependent variable in this experiment is the amount of growth of the corn. It is this because the growth depends on what the scientist did on the corn.
Q11. The variable controlled in this experiment is the amount of sun and water. These two variables never change so this is why it is the control.
Explanation:
