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anygoal [31]
3 years ago
15

Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with t

itration.
Close to the calculated endpoint of a titration ________
At the beginning of a titration _______
Filling the buret with titrant ________
Conditioning the buret with titrant _______
Chemistry
1 answer:
densk [106]3 years ago
3 0

Answer:

Close to the calculated endpoint of a titration - <u>Partially open</u>

At the beginning of a titration - <u>Completely open</u>

Filling the buret with titrant - <u>Completely closed</u>

Conditioning the buret with the titrant - <u>Completely closed</u>

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

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The graph shows the volume of a gaseous product formed during two trials of a reaction. A different concentration of reactant wa
solniwko [45]

A: Trial 1, because the average rate of the reaction is lower.

The rate of reaction is the speed with which reactants are converted into products. It is also the rate at which reactants disappear and products appear.  The higher the rate of reaction, the greater the amount of product formed in a reaction.

If we look at the graph, we will realize that trial 1 produces a lesser amount of product than trial 2. This implies that the  average rate of the reaction in trial 1 is lower than in trial 2.

Lower average rate of reaction implies lower concentration of the reactants since the rate of reaction depends on the concentration of reactants.

Hence trial 1 has a lower concentration of reactants because the average rate of the reaction is lower.

8 0
3 years ago
Read 2 more answers
Part A
den301095 [7]

Answer:

n_{Cl_2}=0.3molCl_2

Explanation:

Hello there!

In this case, according to the given chemical reaction whereas the sodium chloride is in a 2:1 mole ratio with chlorine, the required moles of the later are computed as shown below:

n_{Cl_2}=0.6molNaCl*\frac{1molCl_2}{2molNaCl}

So we cancel out the moles of NaCl to obtain:

n_{Cl_2}=0.3molCl_2

Best regards!

3 0
3 years ago
A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is:
Luden [163]

Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have

64 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 100 X

B)  total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have500 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have20 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 178.88 X

8 0
3 years ago
How many grams of solute are dissolved in 125.0 mL of 5.00 M NaCl (MM = 58.45)?
Nookie1986 [14]

Answer: 36.53g

Explanation:

First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl

Molarity = 5M

MM of NaCl = 58.45

Molarity = Mass conc (g/L) / MM

Mass conc. (g/L) of NaCl = Molarity x MM

= 5 x 58.45 = 292.25g

Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)

From the calculations above,

292.25g of NaCl dissolved in 1L

Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e

Xg of NaCl = 292.25 x 0.125 = 36.53g.

Therefore 36.53g of NaCl will dissolve in 125mL of the solution

5 0
3 years ago
A flame in a sealed container will go out. Explain what you know about combustion reactions to explain why.
klasskru [66]

Answer:

A combustion reaction is a reaction that involves a substance and oxygen reacts at high temperature and produces light energy and thermal energy and generate light and heat.

If a flame is in a sealed container it will go out due to the absence of oxygen to perform a combustion reaction that requires oxygen to produce heat and light. A flame is the result of a combustion reaction which also requires the oxygen to keep on burning.

3 0
3 years ago
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