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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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Allotropes are elements on the periodic table that have more than one crystalline form . There are three forms of the element carbon : Diamond, Graphite, and Fullerenes. Isotopes are atoms of the same element with the same atomic number but have a different mass number.
8 0
3 years ago
Do gases have mass and weight?​
slega [8]

Answer:

yes

Explanation:

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7 0
3 years ago
The standard enthalpy change for the following reaction is 873 kJ at 298 K.
Flura [38]

Answer:  - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

2KCl(s)\rightarrow 2K(s)+Cl_2(g)  \Delta H_1=873kJ

Now we have to determine the value of \Delta H for the following reaction i.e,

K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s) \Delta H_2=?

According to the Hess’s law, if we divide the reaction by half then the \Delta H will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=\frac{1}{2}\times (-873kJ)

\Delta H_2=-436.5kJ

Hence, the value of \Delta H_2 for the reaction is -436.5 kJ.

8 0
3 years ago
Attempt 2
Assoli18 [71]
Lets let our mass equal 3 on alletals and solve using d=m/v equation

Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium
4 0
3 years ago
A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th
Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.  

The mass percent of lithium hydroxide can be calculated with the following equation:  

\% = \frac{m_{LiOH}}{m_{t}} \times 100    (1)

Where:

m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g   (2)  

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.    

\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}

0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}

\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    (3)                                        

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:  

\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    

\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol              

Solving for m_{LiOH}, we have:

m_{LiOH} = 0.082 g

Hence, the percent lithium hydroxide is (eq 1):

\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%  

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

  • brainly.com/question/6992535?referrer=searchResults
  • brainly.com/question/5840377?referrer=searchResults

I hope it helps you!                        

5 0
2 years ago
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