Ask question

Ask question

All categories

2 years ago

7

Based on the law of conservation of mass, if approximately 20g of water reacts in the following equation,

1 answer:

GaryK [48]2 years ago

4 0

Answer: H2O2

Explanation:

First, calculate the moles of H2O2 reacting. In order to do this, we must evaluate the relative molecular mass ( Mr ) of hydrogen peroxide

You might be interested in

Remember KE=M*V2/2…  A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy

pishuonlain [190]

Answer: 4 times as much

Explanation:

4 0

2 years ago

What did you include in your description? Check all that apply

Alexeev081 [22]

I included things
Hope this helps!

7 0

3 years ago

Ill give brainliest! 1. Which item below describes a quick change to

stiks02 [169]

The best answer to go with is b

7 0

2 years ago

Read 2 more answers

: Does increasing temperature increase pressure?

Vesnalui [34]

Yes, an increase in temperature is accompanied by an increase in pressure. Temperature is the measurement of heat present and more heat means more energy. Molecules in hotter temperatures move faster and more often, eventually moving into the gaseous phase. The molecules would fill the container, and the hotter it got the more they would bounce off the walls, pushing outward, increasing the pressure.
I suppose you could measure this with some kind of loosely inflated balloon and subject it to different temperatures and then somehow measure the size/pressure of it.

5 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago