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Nitella [24]
3 years ago
15

A molecule composed of atoms is called _________.

Chemistry
2 answers:
Gelneren [198K]3 years ago
8 0
I think it's element?
Readme [11.4K]3 years ago
3 0
I believe that the answer is element
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For the process 2SO2(g) + O2(g) --> 2SO3(g),
CaHeK987 [17]

Answer:

–187.9 J/K

Explanation:

The equation that relates the three quantities is:

\Delta G = \Delta H - T \Delta S

where

\Delta G is the Gibbs free energy

\Delta H is the change in enthalpy of the reaction

T is the absolute temperature

\Delta S is the change in entropy

In this reaction we have:

ΔS = –187.9 J/K

ΔH = –198.4 kJ = -198,400 J

T = 297.0 K

So the Gibbs free energy is

\Delta G=-198,400-(297.0)(-187.9)=254.2 kJ

However, here we are asked to say what is the entropy of the reaction, which is therefore

ΔS = –187.9 J/K

8 0
3 years ago
g A 0.25 mol sample of HBr is added to a 1.00 L buffer solution consisting of 0.68 M HCN and 0.35 M NaCN. Identify all species t
salantis [7]

Answer:

See explaination for details and diagram

Explanation:

We can say that Chemical reaction is a process in which one or more substances, the reactants, are converted to one or more different substances, the products. Substances are either chemical elements or compounds.

A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products.

See attachment for the buffer of the chemical reaction.

7 0
3 years ago
Please answer asap MUCH NEEDED!!
gulaghasi [49]

Answer:

I think its C I am sorry if I am wrong

3 0
3 years ago
Consider the group a1 element sodium atomic number 11, the group 3a element aluminum atomic number 13, and the group 7a element
Akimi4 [234]
According to there chemical properties
7 0
3 years ago
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0
3 years ago
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