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frozen [14]
3 years ago
13

Describe how to use the periodic table to identify characteristics of elements/atoms.

Chemistry
1 answer:
timofeeve [1]3 years ago
5 0

Answer: The simplest way to use the periodic table to identify an element is by looking for the element's name or elemental symbol. The periodic table can be used to identify an element by looking for the element's atomic number. The atomic number of an element is the number of protons found within the atoms of that element.

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Notice the populations decrease as you go from the bottom of the food chain to the top. Why do you think this is so?
kodGreya [7K]
Exactly what the person above me said
5 0
2 years ago
Read 2 more answers
Someone please do this!!!
Lilit [14]

Answer:

1) Fe = 69.9%

O = 31.1%

2) H = 5.19%

O = 16.5%

N = 28.9%

C = 49.5%

Explanation:

One easy way to do percent compositions is to assume you have 100g of a substance.

1) Lets say we have 100g of Fe2O3.

The total molar mass would be:

= 55.845*2+15.999*3 = 159.687

The molar mass of the Fe2 alone is:

=55.845*2 = 111.69

Thus, the grams of Fe2(out of a 100) could be calculated by multiplying 100g * the molar mass ratio of Fe2 to the whole:

= 100g *\frac{111.69}{159.687} = 69.9431

Which is approximately 69.9%.

We can find the amount of O3 by simply subtracting, as the rest of the compound is made of O3. Thus, the % composition of O3 is 31.1%

You can then do this same process to the next question, getting us the following:

H = 5.19%

O = 16.5%

N = 28.9%

C = 49.5%

4 0
2 years ago
Plz help! I will give five star thanks and I will give brainliest
Harman [31]

Answer:

Conduction

Explanation:

Conduction is the transfer of heat energy from one substance to another or within a substance.

5 0
3 years ago
The melting point of aluminum is 660ºC. At what temperature will aluminum melt if you have 5 grams? 10 grams? Explain your answe
valkas [14]

Answer:

the melting point of aluminum is 660 degrees Celsius.

5 0
2 years ago
Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan
Valentin [98]

Answer : The equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

Explanation :

The given main chemical reaction is:

N_2(g)+2O_2(g)\rightarrow N_2O_4(g);  K

The intermediate reactions are:

(1) N_2O_4(g)\rightarrow 2NO_2(g);  K_a

(2) \frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g);  K_b

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) 2NO_2(g)\rightarrow N_2O_4(g);  \frac{1}{K_a}

(2) N_2(g)+2O_2(g)\rightarrow 2NO_2(g);  (K_b)^2

Thus, the equilibrium constant for this reaction will be:

K=\frac{1}{K_a}\times (K_b)^2

K=\frac{(K_b)^2}{K_a}

Thus, the equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

5 0
2 years ago
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