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4 years ago

Please help me and thank you 14 points

Chemistry

1 answer:

riadik2000 [5.3K]4 years ago

8 0

Sound uses vibrations to transfer :)

Let me know if this helped!

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If 765 mL of a gas at 2.78 atm, has the volume Change to 2.47 L , What is the pressure of the

vampirchik [111]

Answer:

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new pressure

$P_2 = \frac{P_1V_1}{V_2} \\$

From the question

2.47 L = 2470 mL

We have

$P_2 = \frac{765 \times 2.78}{2470} = \frac{2142}{2470} \\ = 0.867206...$

We have the final answer as

Hope this helps you

3 0

3 years ago

Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engin

OlgaM077 [116]

Answer:

(a) 140 F

(b) The temperature rise at the point where the heat is dumped is 2.51 degC

Explanation:

(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:

$\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}$

The temperatures have to be expressed in Rankine (or Kelvin) degrees

$1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F$

(b) The Carnot efficiency of the cycle is

$\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502$

If the efficiency of the plant is 60% of the Carnot efficiency, we have

$\eta=0.6*\eta_{c}=0.6*0.502=0.302$

The heat used in the plant can be calculated as

$Q_i=W/\eta=750MW/0.302=2483MW$

And the heat removed to the heat sink is

$Q_o=Qi-W=2483-750=1733MW$

If the flow of the river is 165 m3/s, the heat per volume in the sink is

$\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3$

Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is

$\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C$

8 0

4 years ago

[98 POINTS HELP ASAP]

netineya [11]

Vacuoles are storage bubbles found in cells. They are found in both animal and plant cells but are much larger in plant cells. Vacuoles might store food or any variety of nutrients a cell might need to survive. They can even store waste products so the rest of the cell is protected from contamination.

5 0

4 years ago

Read 2 more answers

Plz help im being timed

telo118 [61]

Answer:

C=1

A=2

F=3

E=4

Hope this helps good luck

8 0

3 years ago

A gas sample has an original volume of 760 ml when collected at 700 mm and 40oC. If a change is made in the gas temperature whic

alexandr1967 [171]

Answer:

235k

Explanation:

To solve this problem, we will use the general gas equation.

In its simplest form, the general gas equation can be represented as follows:

P1V1/T1 = P2V2/T2

P1 = 700mmHg

V1 = 760ml

T1 = 40 degrees Celsius = 40 + 273.15 = 313.15K ( conversion into standard temperature unit)

P2 = 1.25 atm

We have to covert this into the same unit as the first pressure.

760mmHg = 1atm

x mmHg = 1.25 atm

x = 1.25 * 760 = 950mmHg

V2 = 420ml

T2 = ?

Now, rearranging the equation will yield:

T2 = P2V2T1/P1V1

T2 = 950 * 420 * 313.5/700 * 760

T2 = 235.125K or -38.025 degrees celcius

6 0

3 years ago