Hydrogen, oxygen, and chlorine are some of the __

Since electrons can easily move from one atom to another,

Setler [38]

The correct answer is that last one

3 0

3 years ago

If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of

Ghella [55]

Answer:

The correct answer is - 0.570 grams

Explanation:

The balanced chemical reaction is given by

Cu(NO3)2(aq) + 2NaOH(aq) --------> Cu(OH)2(s) + 2NaNO3(aq)

1.0 mole 2.0 mole 1.0 mole 2.0 mole

number of mol of Cu(OH)2,

n = Molarity * Volume

= $35.0*0.167 = 5.845$ millimoles

As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2

Mass of Cu(OH)2 = number of mol * molar mass

= $97.5*5.845*10^-3$

= 0.570 grams

Thus, the correct answer is - 0.570 grams

8 0

3 years ago

Consider the following balanced equation for the dissociation of barium hydroxide in an aqueous solution.<img src="https://tex.z

Stolb23 [73]

Answer:

a. 0,035M

Explanation:

For the reaction:

Ba(OH)₂(s) ⇄ Ba²⁺(aq) + 2OH⁻(aq)

Ksp is defined as:

Ksp = [Ba²⁺] [OH⁻]²

5x10⁻³ = [Ba²⁺] [OH⁻]²

if is added a solution of 1M of Ba²⁺:

5x10⁻³ = [1M] [OH⁻]²

The addition of barium hydroxide Ba(OH)₂ gives:

[Ba²⁺] = 1M + x

[OH⁻]² = 2x

Replacing:

5x10⁻³ = [1 + x] [2x]²

5x10⁻³ = 4x² + 4x³

Solutions are:

x = -1,00 M

x = -0,036 M

x = 0,035 M → Right answer, there are not negative concentrations.

Thus, solubility is

a. 0,035M

I hope it helps!

4 0

3 years ago

At what atom of the pyridine ring will oxidation and reduction reactions take place?a. 2.b. 4.c. 3.d. 1.

bagirrra123 [75]

Answer:

1

Explanation:

Pyridine is a six membered hetrocyclic aromatic compound. Oxidation of the pyridine ring occurs at position-1.

Just like benzene, pyridine is stable towards mild oxidizing agents, however, pyridine is oxidized by peracetic acid to yield pyridine N-oxide.

7 0

3 years ago

In a typical calorimetry experiment (similar to what you did with the combustion of cheese balls), a small bag of Skittles (10.0

kolezko [41]

This question is incomplete, the complete question is;

In a typical calorimetry experiment (similar to what you did with the combustion of cheese balls), a small bag of Skittles (10.0 g) was combusted and the temperature of the water (1.00 kg) increased by 40.0 °C.

a) How much energy (in calories and Food calories) was released by the combustion of the Skittles

b) How many small bags of skittles would a 140 lb person have to eat in order to equal the amount of food calories burned by walking a marathon?

Marathon = 26.2 miles, average person burns 0.453 food calories/mile pound)

Answer:

a)

Number of calories = 40000 Cal

Number of food calorie = 40 KCal

b)

41.54 bags of skittles are required

Explanation:

Given that;

mass of water = 1kg = 1000g

change in Temp ΔT = 40°C

specific heat of H₂O = 4.84 = 1 Cal

Now

heat absorbed by the water = MSΔT

= 1000gm × 1 cal × 40°C = 40000 Cal

so

Number of calories = 40000 Cal

(since food calories = 1000 small calories)

so

Number of food calorie = 40 KCal

26.2 × 140

= 3668 mile lb

0.453 food calorie is burned for 1 mile pound.

So, amount of food calorie required if the man runs in the marathon will be

0.435 × 3668 mile lb

= 1661.60 food calires.

now

40KCal food calorie is released by 10gm.

So, 1661.60 cal will be released from

⇒ (10gm × 1661.60) / 40 = 415.4 gm

now

1 bag = 10gm,

therefore 415.4 gm = 41.54 bags

∴ 41.54 bags of skittles are required

6 0

3 years ago

Hydrogen, oxygen, and chlorine are some of the ___ elements