Test tube of ammonium chloride (NH4Cl) being heated over a bunsen burner flame. Ammonium chloride decomposes readily when heated, but condenses in the cooler area at the top of the test tube. This is a reversible reaction, where the ammonium chloride decomposes into the gases ammonia (NH3) and hydrogen chloride (HCl).
Answer:
There is no bar graph attached to this question, however, the question can be answered based on the information given in the question.
The answer is A) average level of happiness
Explanation:
In an experiment, the dependent variable is the variable which is measured by the experimenter. It is the variable that responds to changes made to another variable called independent variable.
In the case of this question, it can be determined, even without the bar graph, that the experiment entails how candy allowance affects a child's happiness. Hence, the candy allowance is changed to influence or cause a response in the child's happiness, which is then measured. Therefore, the AVERAGE LEVEL OF HAPPINESS is the dependent variable.
Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have
k =
where, k = rate constant of reaction
Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours
∴ k =
= 0.1467 hours^(-1)
Now, for 1st order reactions: half life =
= 4.723 hours.
Answer:
1.51 X 10^23 ions
Explanation:
The number of ions in 17.1 gm of aluminum sulphate Al2 (SO4)3 =….. [Molar mass of Al2 (SO4)3 = 342 gm]
in one molecule of Al2(SO4)3 there are 5 ions 2 aluminum and 3 sulfate ions
in 2 molecules there are 2X5= 10 ions
in 10 molecules there are 10X5 = 50 ions
molar mass of Al2(SO4)3 = (2 X 26.98) +( 3 X 32.1) + (3 X 4 X 16.0 ) =342.gms = 17.1/342 =0.0500 moles
1 mole =6.02 X 10^23 molecules ( see Avogadros number)
0.0500 moles = 0.0500 X 6.02 X 10^23 molecules =
0.301 X 10^23 molecules = 3.01 X 10^22 molecules
We determined that each molecule of Al2(SO4)3 has 5 ions
so 3.01 X10^22 molecules have 5 X 3.01 X 10^22 ions =
15.05 X 10^22 ions = 1.51 X 10^23 ions
First compute the number of grams of manganese metal required to make 21.7 grams of H2.
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams
Now density = mass/volume
7.43 = 596.75/volume
volume = 596.75/7.43 = 80.31 mL
80.31 mL is the amount of manganese needed.
