(1.) 163.26 g/mol, C11H17N

Beer brewing begins with steeping grains in hot water, releasing the sugars inside. The sugar water is then heated to a boil and

user100 [1]

Answer:

The answers to the question are

a. 166.64 ° F

b. 217990.08 J/hour or 60.55 J/s = 60.55 watts

c. 13.C

Explanation:

a. To solve the question we list out the given variables thus

mass of grain = 16.5 lbs

Temperature of grain = 67 °F

Volume of hot water = 5 gals = ‪0.02273‬ m³

Equilibrium temperature of the mixture = 154 °F

Specific heat capacity of the grain = 0.44 times specific heat capacity of water

Therefore we have

Heat supplied by hot water = heat gained by mixture

Density of the water = 997 kg/m³ which gives

Therefore the mass of the water = (Density of the water) × (Volume of the water) = (997 kg/m³) × ‪(0.02273‬ m³) = 22.66181 kg

Therefore the heat supplied by the water =22.66 kg×1000 g/kg ×4.2 J/g°C×(Tₓ -‪67.78 °C) = ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×(67.78 -‪19.44)

= 95172 × (Tₓ -‪67.78 °C) =668205.7536 J

(Tₓ -‪67.78 °C) = 7.02 from where Tₓ = 74.80 °C = ‪166.64 ° F

The initial temperature (strike temperature) of the hot water = 74.80 °C = 166.64 ° F

b. Where the mixture lost two degrees we have

22.66 kg×1000 g/kg ×4.2 J/g°C×2 °C + ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×2 °C = 217990.08 J therefore the average energy lost per unit time = 217990.08 J/hour or 60.55 J/s

c. To find out how much it cost we have

Heat energy required to raise 5 gallons of water from 110 °F to 166.64 °F we have

22.66 kg×1000 g/kg ×4.2 J/g°C×(74.8 °C-‪43.33 °C) = 2994745.92 J

Energy lost during the heating = 10% = 299474.59 J

Total energy supplied 2994745.92 J + 299474.59 J = 3294220.5 J

Time for heating = 47 minutes, therefore rate of energy consumption = (3294220.5 J)/ (47×60) = 1168.163 Watt 1.168 kW

Cost of energy = 15.C per kilowatt-hour therefore 1.168 kW for 47 minutes will cost

1.168 kW ×47/60×15 = 13.C

therefore it cost 13.C to heat the 5 gallons of tap water initially at 110 ° F to the strike temperature 166.64 °F

6 0

1 year ago

If the kinetic energy of a particle is equal to twice its rest mass, what is the velocity of the particle? Determine if relativi

ivann1987 [24]

Answer:

The velocity of the particle is 2 m/s,

Explanation:

Kinetic energy is defined as energy of the body due to its motion. It is given by :

$K.E=\frac{1}{2}mv^2$

Where :

m = mass of the object

v = velocity of the object

We have , particle with mass m and its kinetic energy is twice its mass.

$K.E=2m$

$2m=\frac{1}{2}mv^2$

$v^2=\frac{4}{1}$

$v=2$

And unit of velocity are m/s , so the velocity of the particle is 2 m/s.

8 0

2 years ago

The information below describes a redox reaction. 3 equations: First: upper C r superscript 3 plus (a q) plus 2 upper C l supers

Marrrta [24]

Answer:

A.

Explanation:

I just took the quiz.

7 0

2 years ago

Diatomic hydrogen gas and diatomic nitrogen gas react spontaneously to form a gaseous product. Give the balanced chemical equati

alex41 [277]

Answer : $NH_{3}$

Explanation : When diatomic hydrogen gas and diatomic nitrogen gas reacts spontaneously it forms ammonia gas as the product.

we will have the following reactions; As all the reactants and products are in the gaseous state the subscript (g) is given to the molecules.

$H_{2}_{(g)} + N_{2}_{(g)} ----> NH_{3}_{(g)}$

But this is not a balanced equation, as on the right side of reactants we have 2 hydrogen atoms and 2 nitrogen atoms which gives the product which has one nitrogen and three hydrogen in the compound state.

So, balancing the equation, we get,

$3H_{2}_{(g)} + N_{2}_{(g)} ----> 2NH_{3}_{(g)}$

We multiplied 2 with the diatomic hydrogen molecule which resulted in production of 2 moles of ammonia as the product.

So, the complete balanced equation is;

$3H_{2}_{(g)} + N_{2}_{(g)} ----> 2NH_{3}_{(g)}$

5 0

2 years ago

Read 2 more answers

3. What is the atomic mass of phosphorous if phosphorous-29 has a percent abundance of 35.5%, phosphorous-30 has a percent abund

daser333 [38]

Answer:

The atomic mass of phosphorus is 29.864 amu.

Explanation:

Given data:

Atomic mass of phosphorus = ?

Percent abundance of P-29 = 35.5%

percent abundance of P-30 = 42.6%

Percent abundance of P-31 = 21.9%

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass / 100

Average atomic mass = (29×35.5)+(30×42.6) + (31×21.9) /100

Average atomic mass = 1029.5 + 1278 + 678.9/ 100

Average atomic mass = 2986.4 / 100

Average atomic mass = 29.864 amu.

The atomic mass of phosphorus is 29.864 amu.

5 0

2 years ago