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Katarina [22]
3 years ago
10

The point in a titration when the added amount of standard reagent is equal to the amount of analyte being titrated.

Chemistry
1 answer:
creativ13 [48]3 years ago
8 0
That point is called as EQUILIBRIUM, and then you have to note that value to compare with others
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A chemist prepares a solution of magnesium chloride MgCl2 by measuring out 49.mg of MgCl2 into a 100.mL volumetric flask and fil
Flura [38]

Answer:  Molarity of Cl^- anions in the chemist's solution is 0.0104 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute

Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.049g}{95g/mol}=5.2\times 10^{-4}moles  

V_s = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

Molarity=\frac{5.2\times 10^{-4}moles\times 1000}{100ml}=5.2\times 10^{-3}mole/L

Therefore, the molarity of solution will be 5.2\times 10^{-3}mole/L

MgCl_2\rightarrow Mg^{2+}+2Cl^-

As 1 mole of MgCl_2 gives 2 moles of Cl^-

Thus 5.2\times 10^{-3}  moles of MgCl_2 gives =\frac{2}{1}\times 5.2\times 10^{-3}=0.0104

Thus the molarity of Cl^- anions in the chemist's solution is 0.0104 M

6 0
3 years ago
Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t
Maurinko [17]

Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN

Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

5 0
3 years ago
For a phase change, H0 = 2 kJ/mol and A S0 = 0.017 kJ/(K•mol). What are
34kurt

Answer:

ΔG = -6.5kJ/mol at 500K

Explanation:

We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:

ΔG = ΔH - TΔS

Computing the values in the problem:

ΔG = ?

ΔH = 2kJ/mol

T = 500K

And ΔS = 0.017kJ/(K•mol)

Replacing:

ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)

ΔG = 2kJ/mol - 8.5kJ/mol

<h3>ΔG = -6.5kJ/mol at 500K</h3>

8 0
3 years ago
Which carboxylic acid is used to prepare the ester shown?
valentinak56 [21]

Answer:

B

Explanation:

The general equation for the reaction of a carboxylic acid with an alkanol to form an ester is shown below;

RCOOH + ROH ------> RCOOR + H2O

Hence; the reactant carboxylic acid can only be the compound (CH3)2-CH-CH2-COOH in accordance with the general reaction equation shown above.

Hence the reaction is;

(CH3)2-CH-CH2-COOH + CH3-CH2OH -------> CH3CH2 OCO-CH2-CH-(CH3)2

3 0
3 years ago
There are several reagents that can be used to effect addition to a double bond including: acid and water, oxymercuration–demerc
ExtremeBDS [4]

The answer & explanation for this question is given in the attachment below.

7 0
3 years ago
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