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2 years ago

Can someone pls pls help me with these questions?!?!

Chemistry

1 answer:

maxonik [38]2 years ago

5 0

Answer:

13) What common form of energy (not including kinetic or potential) is not included in the “Energy Symbols” key that would normally be present in these examples?

1) Sound Energy

2) Magnetic Energy

14)Look carefully at each of the four systems shown above. Knowing what we have discussed about energy conversions, identify (list) at least three different places where this form of energy (sound) should be “produced”.

1) When you open the water tap, in the turbine and the sound of the light bulb.

2) In the light bulb

3) The sound of the kettle, the turbine and when water is boiling.

4) The movement of the wheel of the bicycle, the turbine and the light bulb.

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This is a picture of a large sheet of ice that moves very slowly across the land. This sheet of ice covers the land and is

Ilya [14]

Answer: I think A

Explanation:

3 0

2 years ago

Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow

Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

$\frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)$

$\rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h$

$\frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}$

[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = $\frac{10}{\rho A_{c}}$

$A_{c} = 0.01 m^{2}$

$\frac{dh}{dt}$ + h = 1

$\frac{dh}{dt}$ = 1 - h

$\frac{dh}{1 - h}$ = dt

$\frac{ln(1 - h)}{-1}$ = t + C

Given at t = 0 and V = 0

$A \times h$ = 0

or, h = 0

-ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

C = 0

So, -ln(1 - h) = t

or, t = $ln (\frac{1}{1 - h})$ ........... (1)

(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

t = $ln (\frac{1}{1 - h})$

t = $ln (\frac{1}{1 - 0.6})$

= $ln (\frac{1}{0.4})$

= 0.916 seconds

(b) As maximum height of water level in the tank is achieved at steady state that is, t = $\infty$ .

1 - h = exp (-t)

1 - h = 0

h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

8 0

3 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago

A decrease in the magnitude of velocity is called

IceJOKER [234]

<span>The answer is deceleration. Acceleration is the general term to refer to the change in velocity. Acceleration = change in velocity / change in time. When you want to highlight the fact that the change in velocity is a decrease in the magnitude, you can use the term deceleration, which means that the acceleration is negative.</span>

3 0

3 years ago

Balloon has a volume of 600-ml at temperature of 360 K. If the temperature of

Molodets [167]

Answer:

V₂ ≈416.7 mL

Explanation:

This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.

V₁ / T₁ = V₂ / T₂

where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.

The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,

V₁= 600 mL
T₁= 360 K
T₂= 250 K

Substitute the values into the formula.

600 mL /360 K = V₂ / 250 K

Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.

250 K * 600 mL /360 K = V₂ / 250 K * 250 K

250 K * 600 mL/360 K = V₂

The units of Kelvin cancel, so we are left with the units of mL.

250 * 600 mL/360=V₂
416.666666667 mL= V₂

Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.

416.7 mL ≈V₂

The volume of the balloon at 250 K is approximately 416.7 milliliters.

5 0

2 years ago