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3 years ago

Identify each of the following:

Chemistry

1 answer:

zaharov [31]3 years ago

7 0

Most metallic: Ge

Largest atomic radius: Bi

Highest ionization energy: Cl

Could you mark me as brainliest?

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A sample of a compound contains 160 g of oxygen and 20.2 g of hydrogen. Give the compound's empirical formula.

Maslowich

Amount of oxygen in the compound = 160 g
Amount of oxygen in the compound = 20.2 gm
Mole of oxygen in the compound = 160/16
= 10 moles
Mole of hydrogen in the compound = 20.2/1.01
= 20 moles
Then
The ratio of oxygen to ration of hydrogen = 1:2
So
The empirical formula of the compound is H2O. I hope the answer has come to your help.

3 0

3 years ago

Read 2 more answers

2 QUESTIONS: ANSWER THIS with a proper answer plz. PLZ DO NOT ANSWER JUST FOR MY BRAINLY POINTS!!!!

leonid [27]

Answer:

Si hay algo que sucedió entre ustedes raro o un comportamiento que tuviste con esa persona o que te vio hacer asi como le pudieron decir algo de ti que no le gustara a tu amigo hace que se comporte raro contigo

Explanation:

3 0

3 years ago

5. What type of clouds can be found in the mesosphere?

Masja [62]

Some material from meteors lingers in the mesosphere, causing this layer to have a relatively high concentration of iron and other metal atoms. Very strange, high altitude clouds called "noctilucent clouds" or "polar mesospheric clouds" sometime form in the mesosphere near the poles.

I really hope this helps! I wish you the best of luck!

8 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take

tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = 165°C