Answer:
4 th image
Explanation:
water is being boiled n the 4th image indicating that the change between liquids and gasses is water vapor.
C. 1.0 M Al2O3 would be the best answer
Answer: 483 mL of the cleaning solution are used to clean hospital equipment
Explanation:
The question requires us to calculate the volume, in mL, of solution is used to clean hospital equipment, given that 415g of this solution are used and the specific gravity of the solution is 0.860.
Measurements > Density
Specific gravity is defined as the ratio between the density of a given substance to the density of a reference material, such as water:
The density of a substance is defined as the ratio between the mass and the volume of this substance:
Considering the reference substance as water and its density as 1.00 g/mL, we can determine the density of the substance which specific gravity is 0.860:
Thus, taking water as the reference substance, we can say that the density of the cleaning solution is 0.860 g/mL.
Now that we know the density of the cleaning solution (0.860 g/mL) and the mass of solution that is used to clean hospital equipment (415g), we can calculate the volume of solution that is used to clean the equipment:
Therefore, 483 mL of the cleaning solution are used to clean hospital equipment.
Answer:
2= its color
Explanation:
Transition elements are present in the middle of periodic table. These are d-block elements.
These are 38 elements.
All transition elements have partially filled d orbitals.
They showed color in compound because of d-d transition.
During the d-d transition electron absorbed the energy and emit the reminder energy. The emission is usually in the form of color light.
The color of ion is complementary to the absorbed color.
The transition elements are used as a catalyst in industries such as polymer, petroleum industries.
They are ductile, conduct heat and electricity.
Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol