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Sonbull [250]
3 years ago
5

What is the silver concentration in a solution prepared by mixing 250

Chemistry
1 answer:
kotegsom [21]3 years ago
6 0

Answer:

0.221M

Explanation:

From the question ,

The Molarity of AgNO₂ = 0.310 M

Hence , the concentration of Ag⁺ = 0.301 M

The volume of  AgNO₂  = 250 mL

and,

The Molarity of Sodium chromate = 0.160 M

The volume of Sodium chromate =  100 mL.

As the solution is mixed the final volume becomes ,

250mL +100mL = 350mL

Now, using the formula , to find the final molarity of the mixture ,

M₁V₁ ( initial ) =  M₂V₂ ( final )

substituting the values , in the above equation ,  

0.310M  *  250ml  =  M₂ * 350ml

M₂ = 0.221M

Hence , the concentration of the silver in the final solution = 0.221M

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Answer questions a-c about the Bronsted acid-base reaction below using the identifying letters A-D below each structure. The pKa
GuDViN [60]

Answer:

a  H2CO3 b HCO3- and c H+ and HCO3-

Explanation:

As the pKa value of phenol is more than that of carbonic acid(H2CO3), the carbonic acid will have high Ka value than that of phenol.

  The acid that contain high Ka value act as stong acid.From that point of view H2CO3 is a strong acid than phenol as the Ka value of carbonic acid is greater than that of phenol.

  The conjugate base of H2CO3 is bicarbonate ion(HCO3-)

c The species that predorminates at equilibrium are H+ and HCO3-

8 0
3 years ago
If 12.4 mol of Ne gas occupies 122.8 L, how many mol of Ne would occupy 339.2 L under the same temperature and pressure? Record
8090 [49]

Answer:

3.43×10¹ mol

Explanation:

Given data:

Initial number of  moles = 12.4 mol

Initial volume = 122.8 L

Final number of moles = ?

Final volume = 339.2 L

Solution:

The number of moles and volume are directly proportional to each other at same temperature and pressure.

V₁/n₁  =  V₂/n₂

122.8 L/ 12.4 mol  =  339.2 L / n₂

n₂ = 339.2 L× 12.4 mol  / 122.8 L

n₂ = 4206.08 L.mol /122.8 L

n₂ = 34.3mol

In scientific notation:

3.43×10¹ mol

7 0
3 years ago
Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by the element. rb
devlian [24]

Actually Rb or Rubidium in zero state has the following electron configuration:

<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>

 

However we can see that the ion has a 1 positive charge, which means that it lacks 1 electron, therefore the answer from the choices is:

<span>d. rb+: 1s22s22p63s23p64s23d104p6</span>
3 0
3 years ago
Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemogl
Alex73 [517]

Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

Answer:

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

hemoglobin_{(aq)  +  O_{2(aq)   ---------> hemoglobin.O_{2(aq)

Now, we are also being told to calculate only!, the  initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate  for the reaction can be expressed as :

rate = k [hemoglobin_{(aq)}][O_{2(aq)}]

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ hemoglobin_{(aq) ] = 2 × 10⁻⁹ M

[  O_{2(aq)  ]  =  5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹     ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

7 0
3 years ago
If the value for ΔS is postive, and the value for ΔH is negative, thr reaction will be
dmitriy555 [2]

It follows that the reaction is spontaneous at high temperatures Option A.

<h3>What is ΔS ?</h3>

The term ΔS is referred to as the change in the entropy of the system. Now recall that entropy is defined as the degree of disorderliness in a system. If a system is highly disorderly then it means that it has a high entropy. Also, ΔH has to do with the heat change that accompanies a reaction.

We know that both the entropy and the heat change can both either be positive or negative. Now we know that the equation ΔG = ΔH - TΔS can be used to ascertain whether or not a reaction will be spontaneous. If the result is negative, then the reaction will be spontaneous.

As such, when then it follows that the reaction is spontaneous at high temperatures Option A.

Learn more about spontaneous reaction:brainly.com/question/13790391

#SPJ1

6 0
1 year ago
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