What is HOFBrINCl, and when do you need to use it?

What are the orbital models for these notations?

-BARSIC- [3]

Answer:

use photomath

Explanation:

photomath can give you each step

7 0

3 years ago

Read 2 more answers

The specific heat of iron is 0.46 J/g°C. How many joules will it take to make the temperature of a 150. g bar go up from 25°C to

leva [86]

Answer:

Q = 2.415kj

Explanation:

Q = Mc∆temperature

c = 0.46j/g°c = 0.46kj/kg°c

mass = 150g = 0.15kg

∆temperature = 60 - 25 = 35°c

Q = 0.15 X 0.46 X 35 = 2.415kj

Q = 2.415kj

7 0

3 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

) carbon + oxygen = _____________

scZoUnD [109]

Answer:

d)carbon(iv)oxide/carbon(ii)oxide

e)Calcium,carbon,oxygen

f)carnonhydride

g)Carbonhydrate

h)hydrogen+oxygen

I)Iron sulphide

j)Magnesiumoxide

m)magnesium hydroxide+hydrogen gas

d)carbon dioxide

7 0

4 years ago

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with K

Shalnov [3]

Answer:

The concentration of the unknown acid (HA) is 0.434M

The molar mass of HA is 13.3g/mole

Explanation:

DETERMINATION OF MOLARITY OF THE UNKNOWN ACID

CaVa/CbVb = Na/Nb

From the equation of reaction and at equivalence point, Na = Nb = 1

Therefore, CaVa = CbVb

Va (volume of acid solution) = 20mL = 20/1000 = 0.2L

Cb (concentration of KOH) = 0.715M

Vb (volume of KOH) = 12.15mL

Ca (concentration of acid) = CbVb/Va

Ca = 0.715M × 12.15mL/20mL = 0.434M

DETERMINATION OF MOLAR MASS OF HA

Number of moles of acid = concentration of acid × volume of acid solution in liters = 0.434 × 0.2 = 0.0868mole

Molar mass of HA = mass/number of moles = 1.153g/0.0868mole = 13.3g/mole

7 0

4 years ago