CHEMISTRY HIGH SCHOOL

Leutium-176 has a half-life of 3.85 mc012-1.jpg 1010 years. After 1.155 mc012-2.jpg 1011 years, how much leutium-176 will remain from an original 16.8-g sample?2.10 g
3.00 g
5.56 g
8.40 g

Answers

Answer 1

Answer:

Answer : The amount left of leutium-176 will be, 2.10 g

Solution :

First we have to calculate the rate constant, we use the formula :

$k=(0.693)/(t_(1/2))$

$k=\frac{0.693}{3.85* 10^(10)\text{years}}$

$k=0.18* 10^(-10)\text{years}^(-1)$

Now we have to calculate the amount left of the sample.

Expression for rate law for first order kinetics is given by :

$t=(2.303)/(k)\log(a)/(a-x)$

where,

k = rate constant = $0.18* 10^(-10)\text{years}^(-1)$

t = decay time = $1.155* 10^(11)\text{ years}$

a = initial amount of the sample = 16.8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$1.155* 10^(11)\text{years}=\frac{2.303}{0.18* 10^(-10)\text{years}^(-1)}\log(16.8)/(a-x)$

$a-x=2.10g$

Therefore, the amount left of leutium-176 will be, 2.10 g

Answer 2

Answer:

Answer:

Explanation:

Related Questions

The specific heat of a solution is 4.18 J/(g•°C)and its density is 1.02 g/mL. The solution is formed by combining 25.0 mL of solution A with 25.0 mL of solution B with each solution initially at 21.4°C. The final temperature of the combined solution is 25.3°C. Calculate the heat of reaction, q, assuming no heat loss due to the calorimeter. I got 831 JThe part I cannot figure out is the question afterwards which is If the calorimeter has a heat capacity of 8.20 J/°C and a correction is included to account for the heat absorbed by the calorimeter what is the heat of reaction q
The method used in this experiment is sometimes called the vapor density method. The ideal gas equation can be rearranged to solve for the density of a vapor . Which type or types of measurement is not needed to calculate the density of a gas?
There are three types of plate boundaries. Which choice is NOT a type of plate boundary?A) convergent B) divergent C) thrust D) transform
Calculate the number of atoms of carbon in 3.8 moles of methane (CH4)
Which change occurs during a nuclear fission reaction? (1) Covalent bonds are converted to ionic bonds. (2) Isotopes are converted to isomers. (3) Temperature is converted to mass. (4) Matter is converted to energy

Which of the following is equal to 4 kilograms? A. 4,000 mg
B. 4,000 g
C. 40 g
D. 4,000 cg

Answers

Of course , B. 4,000 g

Which of the following compounds contains both ionic and covalent bonds?Select one:
a. NH3
b. NaNO3
c. CH3OH
d. AlCl3

Answers

Answer:

d. AlCl3

Explanation:

In AlCl₃ each Al³⁺ cation is surrounded by three Cl⁻ anion. Al³⁺ has a small ionic radius giving it a high charge density whereas Cl⁻ has a large ionic radius, low charge, and easily polarizable. Al³⁺ pulls one electron from each of the three Cl⁻ ion thus giving the Al-Cl bond a covalent character.

NaNO3 because Na is a cation allowing it to bond with anions which are NO3. NO3 is a covalent compound because nitrogen and oxygen are sharing electrons when they bond; thus making a covalent bond

Which statement must be true about a chemical system at equilibrium?(1) The forward and reverse reactions stop.
(2) The concentration of reactants and products are equal.
(3) The rate of the forward reaction is equal to the rate of the reverse reaction.
(4) The number of moles of reactants is equal to the number of moles of product.

Answers

(1) is wrong because at equilibrium both the reactant and the product are still being created they are just being created at the same rate meaning that

(3) is the answer. As the concentrations do not change, but the solution/system is still in dynamic equilibrium.

I agree with the answer below (partial credit would go to her/him I guess)
My answer I guess is more visual to explain
1 says --> | <--
2 is assuming because not all chemical system are the same although this could be true in some cases
3 says --> which is true because that's the def of Equilibrium
<--
4-I don't get that one-sorry

How does the degree of disorder of a gas compare to that of a liquid or a solid? please explain your answer. this is a chemistry home work.

Answers

Molecules of gases are spaced far apart from each other compared to liquids and solids. This results in greater disorder in gas molecules. Liquid molecules are spaced slightly closer together than gas molecules, so the disorder is less than gas molecules. Solid molecules are the most tightly packed, and often vibrate in position. This gives solid molecules the least disorder.

When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided thatA) it is a reactant in one intermediate reaction and a catalyst in the other reaction.
B) it is a product in one intermediate reaction and a catalyst in the other reaction.
C) it is a reactant in one intermediate reaction and a product in the other reaction.
D) it is a reactant in both of the intermediate reactions.

Answers

Answer: Option (C) is the correct option.

Explanation:

When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided that it is a reactant in one intermediate reaction and a product in the other reaction.

For example,

$2Al(s) + 2KOH(aq) + 6H_(2)O\rightarrow 2KAl(OH)_(4)(aq)+ 3H_(2)(g)$ ....(1)

$2KAl(OH)_(4)(aq) + 4H_(2)SO_(4)(aq) + 6H_(2)O(l)\rightarrow 2KAl(SO_(4))_(2)(aq)+ 12H_(2)O(s)$ .........(2)

Cancelling the common species in both the equations as follows.

$2Al(s) + 2KOH(aq) + 6H_(2)O\rightarrow \not{2KAl(OH)_(4)(aq)}+ 3H_(2)(g)$

$\not{2KAl(OH)_(4)(aq)}+ 4H_(2)SO_(4)(aq) + 6H_(2)O(l)\rightarrow 2KAl(SO_(4))_(2)(aq)+ 12H_(2)O(s)$

Therefore, on addition we get the equation as follows.

$2Al(s)+ 2KOH(aq) + 4H_(2)SO_(4)(aq) + 22H_(2)O(l)\rightarrow 3H_(2)(g)+ 2KAl(SO_(4))_(2) + 12H_(2)O(s)$

The right answer for the question that is being asked and shown above is that: "C) it is a reactant in one intermediate reaction and a product in the other reaction." When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided that it is a reactant in one intermediate reaction and a product in the other reaction.

You wash dishes for a chemistry laboratory to make extra money for laundry. You earn 12 dollars/hour, and each shift lasts 75 minutes. Your laundry requires 12 quarters/load. How many loads of laundry will each shift pay for if the cost per load rises to 16 quarters?

Answers

You calculate the amount of loads of laundry as follows:

((6 x 0.25)/ load) x 10 loads = 15.00 total cost required for laundry

(6.00 / 60 min) x (75 min/shift) = 7.50 cost / shift

15.00 / (7.50 / shift) = 2 loads of laundry

Hope this answers the question.

Each shift worked can pay for 3.75 loads of laundry if each load costs 16 quarters.

Further Explanation

This problem can be solved using simple dimensional analysis. The steps are:

Sort the given. Identify the conversion factors or equalities that may be used. Identify what is required.
Set up the dimensional analysis ensuring that units cancel out until only the desired unit is left. The equalities in the problem may be used as conversion factors.

STEP 1: Sort first the given in the problem to identify possible conversion factors to be used in the dimensional analysis.

The following equalities are given in the problem:

1 shift = 75 minutes
12 dollars = 1 hour
1 load = 16 quarters
1 dollar = 4 quarters

STEP 2: From these equalities, the following dimensional analysis can be set up:

$no. \ of \ load \ = 1 \ shift * (75 \ min)/(1 \ shift) * {(1 \ hr)/(60 \ min) * (12 \ dollars)/(1 \ hr) * { (4 \ quarters)/(1 \ dollar) * (1 \ load)/(16 \ quarters)$