All categories

3 years ago

STATION 1

Chemistry

1 answer:

grigory [225]3 years ago

3 0

Answer:

B - (C , Al, P, Cl)

Explanation:

How I got this answer was by looking at my periodic table it shows you how much it contains by the Atomic number.

Atomic number on C (Carbon) is- 6

Atomic number on Al (Aluminum) is - 13

Atomic number on P (Phosphorus) is - 15

Atomic number on Cl (Chlorine) is - 17

Now it says least to greatest and the other options are wrong I did the work for you hope this helps :)) I also had this project you didnt ask but the answer for the The Lesson are {B E M S} which as the code numbers are gonna be -7494- Im glad to help if you need more help I will give you the other answers as well :) !

You might be interested in

How many atoms of each element are in the chemical formula P2O5?

tigry1 [53]

2 phosphorus and 5 oxygen

4 0

3 years ago

Read 2 more answers

Read the list of elements given below.

PtichkaEL [24]

Answer:

b All of them are highly reactive nonmetals

3 0

3 years ago

Consider the reaction 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) Using standard thermodynamic data at 298K, calculate the entropy change f

Ulleksa [173]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$4HCl(g)+O_2(g)\rightarrow 2H_2O(g)+2Cl_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cl_2(g))})+(2\times \Delta S^o_{(H_2O(g))})]-[(4\times \Delta S^o_{(HCl(g))})+(1\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(O_2(g))}=205.14J/K.mol\\\Delta So_{(HCl(g))}=186.91J/K.mol\\\Delta S^o_{(Cl_2(g))}=223.07J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (223.07))+(2\times (188.82))]-[(4\times (186.91))+(1\times (205.14))]\\\\\Delta S^o_{rxn}=-129J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-129) J/K = 129 J/K

We are given:

Moles of HCl gas reacted = 2.27 moles

By Stoichiometry of the reaction:

When 4 mole of HCl gas is reacted, the entropy change of the surrounding will be 129 J/K

So, when 2.27 moles of HCl gas is reacted, the entropy change of the surrounding will be = $\frac{129}{4}\times 2.27=73.21J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

3 0

3 years ago

How does the labrador current affect the climate of canada and greenland

max2010maxim [7]

The Labrador current affects the climate of Canada and Greenland because it creates one of the densest concentrations of fog in the world occurs.

3 0

3 years ago

Please help it’s due today I will give brainliest

Thepotemich [5.8K]

1st and 4th options are suitable answers, as these 2 changes are not exactly physical changes as it cant return back to original form and as well as its not cooling, so I feel its 1st and 4th options

7 0

3 years ago