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klemol [59]
3 years ago
10

Which statement indicates that a substance may be an ionic compound?

Chemistry
1 answer:
Free_Kalibri [48]3 years ago
4 0
It can be flattened with force and is made up of crystal lattice.
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What does squeezing do to the particles of snow or ice?​
3241004551 [841]

Answer:

crushes the ice to compaction and satasfation

3 0
3 years ago
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A student described two properties of a substance as shown.
MatroZZZ [7]

Answer:

both are physical properties

Explanation:

A is about it's malleability and B is about it's density. These are physical properties

4 0
3 years ago
The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. How long does it
DedPeter [7]

Answer:

It will take 2378 years for 25% of the C-14 atoms in a sample of the C-14 to decay.

Explanation:

Radioactive decays/reactions always follow a first order reaction dynamic.

Let the initial amount of C-14 atoms be A₀ and the amount of atoms at any time be A

The general expression for rate of reaction for a first order reaction is

(dA/dt) = -kA (Minus sign because it's a rate of reduction)

k = rate constant

(dA/dt) = -kA

(dA/A) = -kdt

 ∫ (dA/A) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from A₀ to A and the right hand side from 0 to t.

We get

In (A/A₀) = -kt

(A/A₀) = e⁻ᵏᵗ

A(t) = A₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus

T(1/2) = (In2)/k

T(1/2) = 5730 years

k = (In 2)/5730 = 0.000120968 = 0.000121 /year.

So, the amount of C-14 atoms left at any time is given as

A(t) = A₀ e⁻⁰•⁰⁰⁰¹²¹ᵗ

How long does it take for 25% of the C-14 atoms in a sample of the C-14 to decay?

When 25% of C-14 atoms in a sample decay, 75% of C-14 atoms in the sample remain.

Hence,

A(t) = 75%

A₀ = 100%

100 = 75 e⁻⁰•⁰⁰⁰¹²¹ᵗ

e⁻⁰•⁰⁰⁰¹²¹ᵗ = (75/100) = 0.75

In e⁻⁰•⁰⁰⁰¹²¹ᵗ = In 0.75 = - 0.28768

-0.000121t = -0.28768

t = (0.28768/0.000121) = 2,377.54 = 2378 years

Hope this Helps!!!

3 0
4 years ago
The oh concentration in a 1.0 x10 3 m ba oh 2 solution is
Lisa [10]
Ba(OH)2 dissociates according to the equation below to yield Barium ions and hydroxide ions.
Ba(OH)2 = Ba²⁺ + 2 OH⁻
The concentration of Ba²⁺ is 1.0 ×10^-3 M
Thus that of OH⁻ ions will be 2× 1.0 ×10^-3 = 2.0 × 10^-3 M
Thus; the answer is 2.0 × 10^-3 M
4 0
3 years ago
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Of the following equilibria, only ________ will shift to the right in response to a decrease in volume. N2 (g) + 3H2 (g) 2NH3 (g
igomit [66]

Answer:

Of the following equilibria, only one will shift to the right in response to a decrease in volume.

N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Decrease the volume

If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.

N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

2 Fe_2O_3 (s) \rightleftharpoons 4 Fe (s) + 3O_2 (g)

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

2 SO_3 (g) \rightleftharpoons 2 SO_2 (g) + O_2 (g)

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

H_2 (g) + Cl_2 (g) \rightleftharpoons 2 HCl (g)

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

2HI (g) \rightleftharpoons H_2 (g) + I_2 (g)

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

8 0
3 years ago
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