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3 years ago

15

In a titration, what is the name of the substance that is being determined?

1 answer:

solmaris [256]3 years ago

8 0

<span>In a titration, the substance that is unknown and being identified is called analyte. A titration is where a known solution or concentration called the titrant is used to identify and measure an unknown substance which is the analyte.</span>

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The image above shows the nucleus of a nitrogen atom. How can the atomic number of nitrogen be determined?

Julli [10]

I think you forgot to attach the diagram along with the question. I am answering the question based on my research and knowledge. "Count the number of protons" is the one among the following choices given in the question by which the <span>atomic number of nitrogen be determined. The correct option among all the options that are given in the question is the second option or option "B".</span>

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3 years ago

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NARA [144]

Answer:

C

Explanation:

the equater is hotter and the poles is cooler

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3 years ago

How many grams of S are in 665 g of SO2

ollegr [7]

Answer: 111g S

Explanation:

SO2 you have = 665 g

SO2 molar mass = 64.066 g/mol

S molar mass = 32.065 g/mol

$665g SO2*\frac{1 molSO2}{64.066gSO2} *\frac{1molS}{3molSO2} * \frac{32.065gS}{1molS} = 110.944... ~ 111gS$

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3 years ago

1. What are the sources of water on Earth?

Lesechka [4]

Rivers, oceans, lakes, ponds

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3 years ago

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The dehydrogenation of benzyl alcohol to make the flavoring agent benzaldehyde is an equilibrium process described by the equati

valkas [14]

Answer:

(a) $p_{C6H5CH2OH}=2.14x10^{-3}atm$

(b) $Dissociation =0.99$

Explanation:

Hello,

(a) In this case, for the given chemical reaction, the law of mass action becomes:

$Kc=\frac{[C6H5CHO][H2]}{[C6H5CH2OH]}$

In such a way, as 1.20 g of benzyl alcohol are placed into a 2.00-L vessel, the initial concentration is:

$[C6H5CH2OH]_0=\frac{1.20g*\frac{1mol}{108.14g} }{2.00L} =5.55x10^{-3}M$

Hence, by writing the law of mass action in terms of the change $x$ due to equilibrium:

$Kc=\frac{(x)(x)}{5.55x10^{-3}-x}=0.558$

Solving for $x$ by using a quadratic equation one obtains:

$x=5.50x10^{-3}M$

Thus, the equilibrium concentration of benzyl alcohol is computed:

$[C6H5CH2OH]_{eq}=5.55x10^{-3}M-5.50x10^{-3}M=5x10^{-5}M$

With that concentration the partial pressure results:

$p_{C6H5CH2OH}=[C6H5CH2OH]_{eq}RT =5x10^{-5}\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*523K \\p_{C6H5CH2OH}=2.14x10^{-3}atm$

(b) Now, the fraction of benzyl alcohol that is dissociated relates its equilibrium concentration with its initial concentration:

$Dissociation=\frac{C6H5CH2OH_{eq}}{C6H5CH2OH_0} =\frac{5.50x10^{-3}M}{5.55x10^{-3}M} =0.99$

Best regards.

3 0

3 years ago

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