Values of Ksp greater than 1.0 infer: 1. The solid phase is relatively insoluble.
2. The solid phase is relatively soluble.
3. The chemical reaction proceeds readily to the right.
4. The chemical reaction proceeds readily to the left.

Answers

Answer 1

Answer:

Values of Ksp greater than 1.0 infer that the solid phase is relatively soluble. The answer is number 2. If Ksp is less than 1 then the solid phase is relatively insoluble.

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Lithium hydroxide is classified as an A. Arrhenius base B. Arrhenius acid C. Alcohol D. Aldehyde

Answers

Lithium hydroxide is classified as an annhenius base since it dissociates into lithium cations and hydroxide an ions and Arrhenius bases are defined as a substance that dissociated into hydroxide anions (as well as cations) when in solution. I hope this helps. Let me know if anything is unclear.

A patient is given 0.050 mg of technetium-99 m (where m means metastable—an unstable but long-lived state), a radioactive isotope with a half-life of about 6.0 hours.How long until the radioactive isotope decays to 1.3×10−2 mg ?

Answers

The correct answer is 9.6h.

As you know, a radioactive isotope's nuclear half-life tells you exactly how much time must pass in order for an initial sample of this isotope to be halved.

Using the formula , A = Ao. $1/2^n$

where , A- final mass after decay

Ao - initial mass

n - the number of half-lives that pass in the given period of time

Now, putting all the values, we get

1.3 × $10^-2$ mg = 0.050 mg × $1/ 2^n$

Take the natural log of both sides of the equation to get,

㏑ $(1.3 . 10^(-2) / 0.050 )$ = ㏑ $((1/2)^(n))$

㏑ $(1.3 . 10^(-2) / 0.050 )$ = n. ln $(1/2)\n$

n = 1.6

Since n represents the number of half-lives that pass in a given period of time, you can say that

$n = t / t _(1/2)$

t= 1.6 × 6 h

t = 9.6h
Hence, it will take 9.6 h until the radioactive isotope decays.

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Final answer:

Using the formula for radioactive decay and the provided half-life of technetium-99m, it can be calculated that it takes approximately 28.5 hours for 0.050 mg of technetium-99m to decay to a quantity of 1.3 x 10^-2 mg.

Explanation:

The decay of a radioactive isotope is an exponential process based on the half-life, which is, in turn, constant for any given isotope. The general formula for the remaining quantity of a radioactive isotope after a given time is given by: N = N0 (0.5) ^(t/t1/2), where (N0) is the initial amount, (N) is the remaining amount, (t) is time, and (t1/2) is the half-life of the isotope. In this case, we are given the initial quantity (N0 = 0.050 mg), the remaining quantity (N = 1.3 x 10^-2 mg), and the half-life (t1/2 = 6.0 hours).

We can solve for time (t) in the equation: N = N0 (0.5) ^(t/t1/2). Plugging in the values, we get 1.3 x 10^-2 = 0.050 x (0.5)^(t/6), and solving for t, we find that it takes approximately 28.5 hours for the technetium-99m to decay to 1.3 x 10^-2 mg.

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If the atomic weight of nitrogen is 14.01, what is the mass of the nitrogen atoms in one mole of cadmium nitrate, Cd(NOÀ)

Answers

The answer is 28.02.

The chemical formula for cadmium nitrate is Cd(NO₃)₂.

There are in total 2 nitrogen atoms in Cd(NO₃)₂. If the atomic weight of nitrogen is 14.01, the mass of two nitrogen atoms in one mole of cadmium nitrate is 28.02:
2 · 14.01 = 28.02

The half-life for the process 238u→206pb is 4.5×109 yr. a mineral sample contains 61.0 mg of 238u and 15.5 mg of 206pb. part a what is the age of the mineral

Answers

The age of the mineral is 8.9×10⁹ years.

We'll begin by calculating the rate constant.

Half-life (t½) = 4.5×10⁹ years

Rate constant (K) =?

$K = (0.693)/(t_½) \n \n K = \frac{0.693}{4 * {10}^(9) } \n \n K = 1.54 * {10}^( - 10) /year \n \n$

Finally, we shall determine the age of the mineral.

Original amount (N₀) = 61.0 mg

Remain remaining (N) = 15.5 mg

Rate constant (K) = 1.54×10¯¹⁰ / year

Time (t) =?

$t = (2.303)/(k) log( (N_0)/(N) ) \n \n t = \frac{2.303}{1.54 * {10}^( - 10) } log( (61)/(15.5)) \n \n t = 8.9 * {10}^(9) years$

Therefore, the mineral is 8.9×10⁹ years old.

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Balance this chemical equation: ___ N2 (g) __O2 (g) → ___ H2O ___ NO3 (aq)

Answers

Answer:

To balance the chemical equation, I'll assume the reaction is carried out under standard conditions. Here's the balanced equation:

2 N2 (g) + 5 O2 (g) → 4 H2O (l) + 4 NO3- (aq)

Explanation:

To balance the number of nitrogen (N) atoms, we need 2 nitrogen molecules (N2) on the left side. This gives us a total of 4 nitrogen atoms on the left side.

To balance the number of oxygen (O) atoms, we need 5 oxygen molecules (O2) on the left side. This gives us a total of 10 oxygen atoms on the left side.

On the right side, we need 4 water molecules (H2O) to balance the hydrogen (H) atoms. This gives us a total of 8 hydrogen atoms on the right side.

Finally, to balance the oxygen (O) atoms on the right side, we need 4 nitrate ions (NO3-) since each nitrate ion contains one nitrogen and three oxygen atoms. This gives us a total of 12 oxygen atoms on the right side.

Now the equation is balanced with 4 nitrogen atoms, 10 oxygen atoms, 8 hydrogen atoms, and 4 nitrate ions on both sides.

The atomic mass of an element can be approximated by adding the number of _______ plus the average number of _______.A. protons; electrons

B. protons; neutrons

C. neutrons; electrons

Answers

Answer: Option (B) is the correct answer.

Explanation:

Atomic mass is defined as the summation of total number of protons and neutrons present in an element.

On the other hand, atomic number means the total number of protons present in an element.

For example, atomic number of sodium is 11 and number of neutrons present in it are 12.

Hence, atomic mass of sodium = no. of protons + no. of neutrons

= (11 + 12) amu

= 23 amu

Therefore, we can conclude that the atomic mass of an element can be approximated by adding the number of protons plus the average number of neutrons.

by adding number of protons and neutrons

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