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nasty-shy [4]
3 years ago
12

Calculate the [ H + ] and pH of a 0.0040 M butanoic acid solution. The K a of butanoic acid is 1.52 × 10 − 5 . Use the method of

successive approximations in your calculations.
Chemistry
1 answer:
stellarik [79]3 years ago
4 0

Answer:

Ka= [H+][But-]/H[Hbut]

Let k moles of H+ be formed

1.52*10^-5=k^2/0.004-k

K^2=0.004*1.52*10^-5-1.52*10^-5k

K^2- 1.52*10^-5k-6.08*10^-8=0

Solving this quadratic equation.

K=7.6*10^-6

PH=-log(7.6*10^-6)=5.2

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The area where the periodic table has the highest electronegativity are Groups 1 and Groups 7 because they are close on giving away an electron and the other is close to becoming stable.  

8 0
3 years ago
What is the solute and solvent in roundup
olga nikolaevna [1]

Answer:

solute is that we disolve in solvent

solvent is in which we dissolve solute

8 0
3 years ago
A fishing line sinker contains 0.650 moles of lead. How many atoms of lead are in the sinker? Show all of your work, including s
34kurt

Answer:

3.91x10²³ atoms of lead

Explanation:

In chemistry, a mole of a substance is defined as 6.022x10²³ particles that could be atoms, molecules, ions, etc.

As you can see, in the problem, you have 0.650moles of lead in a fishing line sinker, the present atoms are:

0.650mol Pb \frac{6.022x10^{23}atoms}{1mol} =<em> 3.91x10²³ atoms of lead</em>

7 0
3 years ago
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
2 years ago
At sea level water blank at 100 degrees celceius
Travka [436]

Boils

Celsius is based on the freezing and boiling points of water.

Water freezes below 0 degrees Celsius (32 degrees Fahrenheit), and it boils above 100 degrees Celsius (212 degrees Fahrenheit).

4 0
2 years ago
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