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Juliette [100K]
3 years ago
11

1. How many atoms of chlorine are present in 1.70x1023 molecules Cl2?​

Chemistry
1 answer:
klasskru [66]3 years ago
6 0

Explanation:

the answer is in the image above

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AICI3(s)+Ca 3N2(s)-AIN(S)+CaCl2(s)
disa [49]

Answer:

2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Chuge%7B%5Cboxed%7B%5Cmathfrak%7BQuestion%3A%7D%7D%7D%5C%20%20%5Ctextless%20%5C%20br%20%2F%5
Kisachek [45]

Explanation:

when hot copper metal reacts with chlorine gas it forms CuCl2 which is yellow in colour.

non metal carbon burns in Oxygen gas they form non metal oxide.

5 0
3 years ago
Read 2 more answers
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
3 years ago
Design an experiment to collect data that supports the claim that a 1.0M NaCl solution is a homogenous mixture.
Alika [10]
The claim is that NaCl mixture is a homogeneous mixture.
Homogeneous mixture means that the components of the mixtures cannot be determined or separated by the naked eye. However, these components can be separated using physical means, such as boiling, evaporation and condensation which will be used in this experiment.

First, we need to prepare one molar solution of NaCl. To do so, we will dilute a mass of 58.44 grams (molar mass of NaCl) in 1 liter of water.
By this, we will have NaCl solution.

We can notice that once the NaCl is diluted in water, all what you can see is a clear solution. You cannot see the separate particles of NaCl in water.
..............> observation I

Now, we will heat this solution until it boils and water starts evaporating. We will place a cold surface above the steam coming out from the boiling solution.

What we will observe is that when all the water evaporates, we can see white precipitate of NaCl in the bottom of the container. Examining the cold surface placed above the steam, we can see that the water has condensed on this surface.
.........>observation II

Based on this, we managed to use boiling, evaporation and condensation (physical methods) to restore the components of the solution separately.
.............>conclusion

Based on observation I, observation II and the conclusion. we were able to prove that NaCl solution is a homogeneous mixture.
7 0
4 years ago
Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s
Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

                        N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where N is amount of radioactive nuclide after t time, N_{0} is initial amount of radioactive nuclide and t_{\frac{1}{2}} is half life of radioactive nuclide

Here N_{0}=4.60 mg, t=4.1\times 10^{9}yr and t_{\frac{1}{2}}=4.5\times 10^{9}yr

So,N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0
3 years ago
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