All categories

3 years ago

Question 13 (1 point)

Chemistry

1 answer:

crimeas [40]3 years ago

7 0

Its a compound because they come together

You might be interested in

What pressure will be exerted by 0.450 mol of a gas at 25°C if it is contained in a 0.650-L vessel?

iVinArrow [24]

Hey there!

The answer as well as the explanation is in the image attached. Let me know if there's anything you're unable to see.

Hope this helps!

5 0

3 years ago

# of protons = 48 Cadmium Titanium Oxygen Krypton

Dafna11 [192]

Answer:

Cadmium +p= 48

Titanium p=22

Oxygen p=8

Krypton p=36

So the answer is Cadmium which has #48 protons

Explanation:

7 0

2 years ago

1 point If the boy is pushing with 50N of force and the static friction resistance is 70N of force, what will happen? *

solniwko [45]

Nothing, he shouldn’t be able to move it. Think about it like this say you try really hard to push something that is 5,000 pounds and you push as hard as you can. Well you can’t move it bc it weighs more than you can push. I’m sure their is a equation you can use to see how much you can push (body weight=force?)

6 0

3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

2 years ago

The temperature and number of moles of a gas are held constant. Which of the following is true for the pressure of the gas?

Taya2010 [7]

Answer:

Pressure is inversely proportional to the volume of gas.

Explanation:

According to Boyle's law,

The volume of given amount of gas is inversely proportional to the pressure applied on gas at constant volume and number of moles of gas.

Mathematical expression:

P ∝ 1/ V

P = K/V

PV = K

when volume is changed from V1 to V2 and pressure from P1 to P2 then expression will be.

P1V1 = K P2V2 = K

P1V1 = P2V2

3 0

3 years ago