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Varvara68 [4.7K]
3 years ago
11

100 points! Please help! Will give brainliest!!

Chemistry
1 answer:
patriot [66]3 years ago
3 0
1- change of state (boiling)
2-increase in acidity
3-methylxanthines (caffeine) theobromine and theophylline
4- lemon juice
You might be interested in
Metric conversions.<br> Please help ASAP.
lesya692 [45]

Answer:

14. 13.2cg = 1.32dg

15. 3.8m = 0.0038km

16. 24.8L = 24800mL

17. 0.87kL = 870L

18. 26.01cm = 0.0002601km

19. 0.001hm = 10cm

Explanation:

14. 13.2/10 = 1.32

15. 38/1000 = 0.0038

16. 24.8(1000) = 24,800

17. 0.87(1000) = 870

18. 26.01/100000 = 0.0002601

19. 0.001hm(10000) = 10

An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.

7 0
3 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of
jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = 2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x

2330.35x=1048815\\x=450.068g

Hence, there was 450.068g of water in the pot.

8 0
3 years ago
What is meant by the ground state of an atom? What is meant by the ground state of an atom? All of the quantum numbers have thei
ANTONII [103]

Answer: Option (a) is the correct answer.

Explanation:

We know that ground state means in total electrons acquired by the element are present in their lowest energy level.

We calculate the azimutal numbers as follows.

                      n = l + 1

where,     n = principle quantum number  

                l = azimuthal quantum number

Values of n can be 1, 2, 3, 4 and so on. Whereas the values of l can be 0, 1, 2, 3, and so on.

Also, "m" is known as magnetic quantum number whose values can be equal to -l and +l.  

So, when n = 1 then l = 0 and m = 0.

When n = 2 then l = 1 and values of m will be equal to -1, 0, +1. As it is given that the magnetic quantum number ml = -1. Hence, it is only possible when n = 2.

And, n = 1 is the lowest energy level. Therefore, we can conclude that the ground state of an atom means all of the quantum numbers have their lowest values (n = 1,ℓ = mℓ = 0).

5 0
3 years ago
The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J
Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

<em>A = 1,13x10¹⁰</em>

<em></em>

I hope it helps!

6 0
3 years ago
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