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erastova [34]
2 years ago
12

The iodide (I2) content of a commercial mineral water was measured by two methods that produced wildly different results.6 Metho

d A found 0.23 milligrams of I2 per liter (mg/L) and method B found 0.009 mg/L. When Mn21 was added to the water, the I2 content found by method A increased each time that more Mn21 was added, but results from method B were unchanged. Which of the Terms to Understand describes what is occurring in these measure- ments? Which result is more reliable?
Chemistry
1 answer:
hammer [34]2 years ago
3 0

Answer:

Results from method B is more reliable than method A.

Explanation:

The two method that are used for the analysis produced different results. The first method that is method A  gives higher value of the iodine content than the method B.

When $Mn^{2+}$ was added to water, method A showed an increased in the iodine content and it increases with the increase in the amount of $Mn^{2+}$ .

Where as in the method B, there is no change in the results. Therefore the measurements provided by the method A shows an inference of $Mn^{2+}$ ion.

The measurement of the iodine content is affected by the presence of the ion $Mn^{2+}$ in water.

Since in method B there is no change in measurement, it is independent of the presence $Mn^{2+}$ ion in water.

As higher iodine content is given by method A, so $Mn^{2+}$  ion must be present in original water that must be interfering the measurement. Hence, method B is more reliable.

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If 10.62 mL of a standard 0.3330 M KOH solution reacts with 98.20 mL of CH3COOH solution, what is the molarity of the acid solut
Nikolay [14]

Answer:

0.036 M of CH_{3} COOH

Explanation:

It is an example of acid-base neutralization reaction.

KOH  + CH_{3} COOH  ----> CH_{3} COO^{-} K^{+}   +   H_{2}O

Base           Acid                           Salt                                    

When two component react then the number of moles of both the component should be same, therefore the number of moles and acids and bases should be the same in the following .

Molarity= \frac{\textrm{No. of Moles}}{\textrm{Volume of the Particular Solution}}

No.of moles= Molarity × Volume of the Particular Solution

Therefore,

M_{1}V_{1} =M_{2}V_{2}------------------------------(1)

where

M_{1}= Molarity of Acid

V_{1}= Volume of Acid

M_{2}= Molarity of Base

V_{2}= Volume of Base

M_{1}=0.3330 M

V_{1}=10.62 mL

V_{2}=98.2 mL

M_{2}=??(in M)

Plugging in Equation 1,

0.3330 × 10.62 =M_{2}  × 98.2  

M_{2}=\frac{0.3330*10.62}{98.2}

M_{2}=0.036 M

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Hi I’m kurama have some of my points!
Burka [1]
✨Tysm for the points kurama :)✨
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Ethyl acetate reacts with H2, in the presence of a catalyst to yield ethyl alcohol: C4H8O2 + 2H2--> 2C2H6O. A. How many grams
Anni [7]
The balanced chemical reaction is:

<span>C4H8O2 + 2H2--> 2C2H6O

</span>A. How many grams of ethyl alcohol are produced by reaction of 2.7 mol of ethyl acetate with H2? 2.7 mol C4H8O2 ( 2 mol C2H6O / 1 mol C4H8O2) (46.07 g / 1 mol) = 248.78 g ethyl alcohol

B. How many grams of ethyl alcohol are produced by reaction of 13.0g of ethyl acetate with H2?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol C2H6O / 1 mol C4H8O2) (<span>46.07 g / 1 mol) = 13.59 g ethyl alcohol</span>

C. How many grams of H2 are needed to react with 13.0g of ethyl acetate?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol H2 / 1 mol C4H8O2) (2.02 g / 1 mol) = 0.5961 g H2
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What is the solubility product (or ion-product), Ksp, expression at equilibrium for calcium phosphate, Ca3(PO4)2?
Gnoma [55]

Answer : The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2

Explanation :

Solubility product : It is defined as the product of the concentration of the ions  that present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only.

The solubility product constant is represented as, K__{sp}.

The dissociation of calcium phosphate is written as:

Ca_3(PO_4)_2\rightleftharpoons 3Ca^{2+}+2PO_4^{3-}

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2

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