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andreev551 [17]

3 years ago

13

How many significant figures? A) 1234500_____ B) 1230.0_ C) 0.10220 D) 34560000 E) 0.0000302010___ F) 0.0205

0______ G) 1.250 x 105_____

1 answer:

Setler79 [48]3 years ago

4 0

Answer:

A) 5 sig figs

B) 4 sig figs

C) 4 sig figs

D) 4 sig figs

E) 5 sig figs

F) 3 sig figs

G) 5 sig figs

Explanation:

work for G- 1.250 x 105= 131.25

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Colligative property definition chemistry

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For the reaction shown, calculate how many moles of NO2 form when each amount of reactant completely reacts.

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Answer:- 2.4 mol $NO_2$ .

Solution:- It asks to calculate the moles of $NO_2$ formed when 1.2 moles of $N_2O_5$ are reacted.

There is 2:4 mol ratio between $N_2O_5$ and $NO_2$ . So, the moles of reactant are multiplied by the mol ratio to get the moles of $NO_2$ . The calculations are shown below:

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3 years ago

Read 2 more answers

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

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3 years ago

Suppose you needed a 0.0250 M sodium thiosulfate solution to conduct 6 titrations. Explain how you would make up this solution u

AysviL [449]

Explanation:

According to the law of dilution,

$M_{1}V_{1} = M_{2}V_{2}$

The given data is as follows.

$M_{1}$ = 0.4782, $V_{1}$ = ?

$M_{2}$ = 0.025 , $V_{2}$ = 250 mL

Hence, we will calculate the value of $V_{1}$ as follows.

$V_{1} = \frac{M_{2} \times V_{2}}{M_{1}}$

= $\frac{0.025 \times 250}{0.4782}$

= 13.07

Thus, we can conclude that we need 13.07 mL 0.4782 M sodium thiosulfate solution using pipette.

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