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3 years ago

IF YOU GIVE A LINK OR DON"T ACTUALLY ANSWER THIS QUESTION RIGHT JUST FOR POINTS I WILL REPORT YOUR ANSWER AND YOU

Chemistry

2 answers:

FromTheMoon [43]3 years ago

7 0

Karen you get back to the kitchen please and give me food and then come on me

frozen [14]3 years ago

4 0

Answer:

let's be honest, chocolate may be a solid but it's both liquid and solid and the reason why it smells is bc the chocolate bar releases gas or some smell or it has a reaction. Many solid release gases.... or a smell

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I WILL GIVE BRAINLIEST TO BEST AND MOST THOROUGH ANSWER!

andrew11 [14]

At 100 feet, the diver is under about 4 atmospheres pressure. If she is free diving, her lungs will be compressed to about 1/4 their size on the surface (with some movement of the major abdominal organs). If she is scuba diving, the air which she is breathing is also at 4 atmospheres and there is no problem. (The non-gas spaces in the body are not-compressible and are unaffected.) The only problems she has to concern herself with are the beginnings to nitrogen narcosis and the nitrogen which is dissolving (Henry's law) into her body tissues. On the way up, she also has to remember that the air in her lungs will expand by a factor of 4 and she better exhale! Hope this helps you

7 0

3 years ago

Predict the products of La(s) + O2(aq) ->

mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

Hello there!

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

I hope this helps you to understand better. Enjoy your studies

3 0

3 years ago

Please, can i get some help with these question? Can anyone please answer what exactly i should write?

pav-90 [236]

Use photomath. I would really recommend it! Hope this helps!

8 0

3 years ago

Read 2 more answers

If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?

kodGreya [7K]

Answer:

$[Cu^{2+}]=0.041 M$

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

$[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}$

$[Cu^{2+}]=0.041 M$

Best regards!

6 0

3 years ago

If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield

Zigmanuir [339]

<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

$ActualYield=37.6g\\TheoreticalYield=112.8g$

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

$PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)$

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0

4 years ago

Read 2 more answers