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marshall27 [118]
2 years ago
7

Water can take on three forms which include ice, water, and water vapor. Which states of matter match these three forms of water

? Write your answer in complete sentences!
Chemistry
2 answers:
Tanya [424]2 years ago
8 0

Answer:

Water as itself can match the state of liquid, ice can match the state of a solid, and water vapor can match the state of gas.

Explanation:

grin007 [14]2 years ago
4 0
The three forms are ice is a solid, water vapor is a gas, and water itself is a liquid
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What happens when a catalyst is added to a chemical reaction that is already in progress?
Elanso [62]

Explanation:

when the catalyst is added the reaction with the lower activation energy occur

3 0
3 years ago
A bathtub is filled with 100 L of water. A cat jumps in and the volume rises to 125 L. If the cat has a mass of
kirza4 [7]

Answer:

<h2>The answer is 3.2 g/L</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of cat = 400 g

volume = final volume of water - initial volume of water

volume = 125 - 100 = 25 L

The density of the cat is

density =  \frac{400}{125}  =  \frac{16}{5}  \\

We have the final answer as

<h3>3.2 g/L</h3>

Hope this helps you

3 0
3 years ago
The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−
barxatty [35]

Answer:

The reaction will be  non spontaneous at these concentrations.

Explanation:

AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)

Expression for an equilibrium constant K_c:

K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]

Solubility product of the reaction:

K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}

Reaction between Gibb's free energy and equilibrium constant if given as:

\Delta G^o=-2.303\times R\times T\times \log K_c

\Delta G^o=-2.303\times R\times T\times \log K_{sp}

\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]

\Delta G^o=69,117.84 J/mol=69.117 kJ/mol

Gibb's free energy when concentration [Ag^+] = 1.0\times 10^{-2} M and [Br^-] = 1.0\times 10^{-3} M

Reaction quotient of an equilibrium = Q

Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}

\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)

\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])

\Delta G=40.588 kJ/mol

  • For reaction to spontaneous reaction:  \Delta G.
  • For reaction to non spontaneous reaction:  \Delta G>0.

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0
3 years ago
Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI, and the rate constant is 9.7×10−6M−
Lady bird [3.3K]

Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 9.7\times 10^{-6}M^{-1}s^{-1}

t = time taken  = 5.00 days

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.110 M

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

9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)

[A]=0.109M

Hence, the molarity after a reaction time of 5.00 days is, 0.109 M

8 0
3 years ago
Which one of the following will change the value of an equilibrium constant?
Ad libitum [116K]

Answer:

(E) changing temperature

Explanation:

Consider the following reversible balanced reaction:

aA+bB⇋cC+dD

If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:

Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)

where:

[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1

There are some important things to remember when calculating Kc:

-  <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes

- Pure solids and liquids, including solvents, are not considered for equilibrium expression.

- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc

8 0
3 years ago
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