Chemistry High School

Which chemical symbol represents an element?
O H2O
O C20
O H₂CO3
O CO

Answers

Answer 1

Answer:

C20

Explanation:

Element is a pure substance which is composed of atoms of similar elements. It can not be decomposed into simpler constituents using chemical reactions. Example: Carbon

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass. It can be decomposed into simpler constituents using chemical reactions.

Example: Carbon monoxide.

Thus chemical symbol represents an element is C20

Pretty sure this is what you're looking for.

Related Questions

How many DIFFERENT ELEMENTS are in this compound?
H2SO4

Answers

Answer:

3 DIFFERENT ELEMENTS

Explanation:

H - Hydrogen

S- Sulfur

O- Oxygen

Therefore there are 3 elements

A solution is prepared by adding 91.9 g of ethanol, ch3ch2oh, to 481.0 ml of water. the final volume of the solution is 597.5 ml. assuming the density of water is 1.0 g/ml and the density of ethanol is 0.789 g/ml, calculate the molality of the ethanol in the solution.

Answers

Answer:

17.51M

Explanation:

first we make out what we've been given .

Mass of ethanol= 91.9g

volume of water = 481.0ml

volume of solution= 597.5ml

density of water = 1.0g/ml

density of ethanol= 0.789g/ml.

firstly find the volume of ethanol since the total volume of the solution is 597.5ml and we've been given the volume of water which is 481.0ml , we subtract the volume of water from the volume of the solution , to find out the volume of ethanol .

volume e =597.5ml - 481.0ml

= 116.5ml

Then we find the molecular mass of ethanol , which is 45.04g/mol .

then we find the number of moles of ethanol .

n=mass /molar mass

=91.9g/ 45.04g/mol

=2.04mols .

Then convert the volume from milliliters to liters , by dividing by a 1000 .

v= 116.5/1000

=0.1165L

since we have our number of moles and volume , we can now find the molarity . which is

M=n/v

= 2.04mols /0.1165L

= 17.51M .

PLEASE SOMEBODY EXPLAIN THIS :(((

Answers

Answer:

[tex] \boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}[/tex]

Explanation:

In Analytical Chemistry chromatography is widely used for the separation of samples.

In thin layer chromatography, the mixture of components are separated on the basis of their polarity. The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.Consider the mixture we have taken consist of two samples having large polar difference.Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.

In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.

Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.

Formula to calculate Rf value is

[tex] \sf R_f \: value = \frac{distance \: moved \: by \: sample}{distance \: moved \: by \: solvent} [/tex]

Now, solving for Rf value of sample 1

Given:

Distance moved by sample 1 = 11 mm

Distance movedby solvent = 35 mm

To find:

Rf value of sample 1 = ?

Solution:

Substituting the given data in above formula,

[tex] \small \sf R_f \: value = \frac{distance \: moved \: by \: sample \: 1}{distance \: moved \: by \: solvent} \\ \small \sf R_f \: value = \cancel\frac{11 \: mm}{35 \: mm} = 0.3142[/tex]

[tex] \small \boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}[/tex]

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Suppose 3.5g of calcium chloride reacts with 6.39g of silver nitrate.

How many moles of calcium nitrate would be produced?

Balanced Equation =
Limiting Reactant =
Excess Reactant =

Answers

CaCl₂+2AgNO₃⇒Ca(NO₃)₂+2AgCl

mole CaCl₂: 3.5 g : 111 g/mole = 0.032

mole AgNO₃: 6.39 g : 170 g/mole = 0.038

mole: coefficient

CaCl₂: 0.032 : 1 = 0.032

AgNO₃: 0.038 : 2 = 0.019

AgNO₃ smaller= limiting reactant, then CaCl₂ = excess reactant

A gas has an initial volume of 5. 35l. What is the final volume if 588j of work is done on the gas by a constant external pressure of 2. 00atm?

Answers

Work done is the exchange of heat from the system and the surrounding. The final volume of the gas in the system is 2.45 L.

What is work done?

Work done is the product of the pressure and the change in the volume of the gas in the system. The formula for work done is given as,

[tex]\rm w = -P\Delta V[/tex]

Given,

Initial volume = 5.35 L

Work done = 588 J

Pressure = 2.00 atm

Work done in joules is converted to L atm as, 588 joules = 5.80 Litres-atmosphere

Substituting values in the above equation:

[tex]\begin{aligned}\rm w &=\rm -P(V_{f} - V_{i})\\\\5.80 &= \rm - 2 (V_{f} - 5.35))\\\\5.80 - 10.7 &= \rm -2V\\\\&= 2.45\;\rm L\end{aligned}[/tex]

Therefore, 2.45 L is the initial volume of the gas.

Learn more about work done here:

https://brainly.com/question/25086088

#SPJ4

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