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Varvara68 [4.7K]
3 years ago
9

The difference between air temperatures over water and over

Chemistry
1 answer:
lakkis [162]3 years ago
5 0
The answer is A
Trust. Me
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A buffer consists of 0.45 M CH3COOH (acetic acid) and 0.35 M CH3COONa. The Ka of acetic acid is 1.8 x 10-5 . a) Calculate the pH
Zigmanuir [339]

Answer:

A) pH of Buffer solution = 4.59

B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original    buffer solution = 4.65

Explanation:

This  is the Henderson-Hasselbalch Equation:

 pH = pKa + log\frac{[conjugate base]}{[acid]}

to calculate the pH of the following Buffer solutions.

8 0
3 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
what is the density of mercury if 205 mL has a mass of 2790 g; round your answer to the nearest 0.1 g/mL
kherson [118]
D = m / V

D = 2790 g / 205 mL

D = 13.60 g/mL
4 0
3 years ago
Which of the following is the correct formula used for calculating Speed?
MAXImum [283]

Answer:

A- Speed = distance/time

Explanation:

Would make sense and that's what I remember

3 0
3 years ago
Metals and their ore​
Maksim231197 [3]

Answer:

Zinc

Zinc blende

(ZnS)

-----------------

Lead

Galena

(PbS)

-----------------

Mercury

Cinnabar

(HgS)

----------------

Copper

Copper glance

(Cu2S)

-------------------

Explanation:

Had to learn this for a test

Also, I plead the 5th

8 0
3 years ago
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