Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
Answer:
Described by a redox reaction below
Explanation:
Iron(III) oxide is an ionic compound, since it consists of a metal, iron, and a nonmetal, oxygen.
Ionic compounds are formed when metals lose their valence electrons in order to have an octet in their previous shell and donate them to nonmetal atoms, so that nonmetals fill their outer shell to have an octet.
As a result, positive ions (cations) and negative ions (anions) are formed. When iron reacts with oxygen, the following reaction takes place:
This is a redox (oxidation–reduction) reaction, since we have electron loss and gain. Four iron atoms lose a total of 12 electrons to obtain a +3 charge in the final compound, while 3 oxygen molecules gain these 12 electrons to become 6 oxide anions with a -2 charge.
The d-block elements are found in groups 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 of the periodic table.
So your answer is A) Transition metals
Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
