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ANTONII [103]
3 years ago
6

Which of these products is made from an animal product?

Chemistry
1 answer:
RUDIKE [14]3 years ago
6 0
I’m not sure but I think it’s B

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I have this chemistry question and I’m really struggling to answer it. Can someone please help me figure out the steps & the
Digiron [165]

Answer: 133,333.333 meters.

Explanation:

I forgot how to explain this looool.

7 0
3 years ago
A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give
8_murik_8 [283]

<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m

<u>Explanation:</u>

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}

Where,

m_{solute} = Given mass of solute (KOH) = 3.301 g

M_{solute} = Molar mass of solute (KOH) = 56.1 g/mol

W_{solvent} = Mass of solvent = 96.699 g

Putting values in above equation, we get:

\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m

Hence, the molality of potassium hydroxide solution is 0.608 m

4 0
3 years ago
100 points! Report on an element
Luden [163]

Hydrogen  H weight: 81

Non-metal

Hydrogen is the simplest element; an atom consists of only one proton and one electron. It is also the most plentiful element in the universe. Despite its simplicity and abundance, hydrogen doesn't occur naturally as a gas on the Earth--it is always combined with other elements.

period 1 group 1

Hydrogen is easily the most abundant element in the universe. It is found in the sun and most of the stars, and the planet Jupiter is composed mostly of hydrogen. On Earth, hydrogen is found in the greatest quantities as water.

6 0
3 years ago
Read 2 more answers
Logging trees is often controversial. Why would some people support logging and some people not support logging?
Harlamova29_29 [7]
It seems more and more there are fewer conservation organizations who speak for the forest, and more that speak for the timber industry. Witness several recent commentaries in Oregon papers that are by no means unique. I’ve seen similar themes from other conservation groups across the West in recent years.

Many conservation groups have uncritically adopted views that support more logging of our public lands based upon increasingly disputed ideas about forest health and fire ecology, as well as the age-old bias against natural processes like wildfire and beetles.

For instance, an article in the Portland Oregonian quotes Oregon Wild’s executive director Sean Stevens bemoaning the closure of a timber mill in John Day Oregon. Stevens said: “Loss of the 29-year-old Malheur Lumber Co. mill would be ‘a sad turn of events’” Surprisingly, Oregon Wild is readily supporting federal subsidies to promote more logging on the Malheur National Forest to sustain the mill.
7 0
3 years ago
Read 2 more answers
The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
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