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Lana71 [14]
2 years ago
5

So I’m in 6th grade and this is a chart. I need the # of molecules, # of atoms in each element and the total # of atoms 2H₂SO

Chemistry
1 answer:
xxTIMURxx [149]2 years ago
4 0

Answer:

1.) 2 molecules

2.) 4 H atoms, 2 S atoms, 2 O atoms

3.) 8 total atoms

Explanation:

1.) 2 H₂SO

--------> The molecule is all of the elements bonded together (H₂SO). Because there is a coefficient of 2 in front of the molecule, this means that there are 2 molecules present.

2.) 2 H₂S₁O₁

---------> The number of atoms of each element is determined by the subscripts and the coefficients. In one molecule, there is one oxygen, one sulfur, and 2 hydrogen atoms (denoted by subscripts). Because there are two molecules, these values are doubled.

3.)

--------> The total number of atoms is calculated by adding the number of atoms of each element. If between 2 molecules there are 4 hydrogens, 2 sulfurs, and 2 oxygens, combined there would be a total of 8 atoms.

You might be interested in
Glucose (C6H12O6)(C6H12O6) can be fermented to yield ethanol (CH3CH2OH)(CH3CH2OH) and carbon dioxide (CO2).
zhenek [66]

Answer: a) 49.8 gram

b) 47.0 %

Explanation:

First we have to calculate the moles of glucose

\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{97.5g}{180.15g/mole}=0.54moles

The balanced chemical reaction will be,

C_6H_{12}O_6\rightarrow 2CH_3CH_2OH+2CO_2

From the balanced reaction, we conclude that

As,1 mole of glucose produce = 2 moles of ethanol

So, 0.54 moles of glucose will produce =  \frac{2}{1}\times 0.54=1.08 mole of ethanol

Now we have to calculate the mass of ethanol produced

\text{Mass of ethanol}=\text{Moles of ethanol}\times \text{Molar mass of ethanol}

\text{Mass of ethanol}=(1.08mole)\times (46.08g/mole)=49.8g

Now we have to calculate the percent yield of ethanol

\%\text{ yield of ethanol}=\frac{\text{Actual yield}}{\text{Theoretical yield }}\times 100=\frac{23.4g}{49.8g}\times 100=47.0\%

Therefore, the percent yield is 47.0 %

7 0
3 years ago
In a chemical equation, the symbol that takes the place of the word yields is an) Example: HCI + KOH-KCI + H2O
Masteriza [31]

Answer:

Arrow.

Explanation:

Hello!

In this case, since chemical reactions are characterized by the presence of some species at the left on an arrow representing the reactants and at the right the products, we understand that the arrow, separating reactants and products is understood as "produces" or in very chemical words "yields"; therefore, the symbol that takes the place of the word yields is an arrow.

Best regards!

4 0
3 years ago
JJ Thomson's cathode-ray tube experiment showed which of the following?
victus00 [196]
The answer would be D
6 0
4 years ago
Decide which element probably has a boiling point most and least similar to the boiling point of cesium.
natka813 [3]

Answer:

Take a look at the attachment below

Explanation:

Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.

Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.

<em>Hope that helps!</em>

5 0
3 years ago
5. There is water on the pan of the scale as you measure the mass of a mineral. If you were to
fgiga [73]

The density is calculated with the following formula:

density = mass / volume

If there is water on the pan of scale, you will read the mass of the mineral plus mass of water.

Using this value to calculate the density (ignoring the fact that you had water) you will obtain a higher density of the mineral than the real one.

3 0
4 years ago
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