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Jlenok [28]
3 years ago
11

What do the arrows and coefficients used by chemists in equations communicate?

Chemistry
2 answers:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

In a chemical reaction, arrows separate reactants from products and points to where the reaction is going. On the other hand, the coefficients detail the amount of the components of the chemical reaction.

Explanation:

The way to represent a chemical reaction is through the chemical equation. To write this chemical equation, the following rules must be followed: The reactants are written on the left, and the products on the right, both, separated by an arrow that indicates the direction of the reaction.

The chemical equation must be balanced, leaving the same number of atoms of each of the elements. To balance a chemical reaction, numbers known as coefficients are placed in front of it, showing the relative number of atoms entering a reaction.

zmey [24]3 years ago
3 0
The arrows used by chemists means "react to produce" or "yield", such that when two substances react with each other they yield products, represented or shown by an arrow. The coefficients on the other hand, describe the lowest whole number ratio of the amounts of all of the reactants and products.
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Simple question. How many groups and how many periods are on the periodic table? Which directions do they go?
stepan [7]
There is 18 groups in the periodic table..  periodic tables go from left to right. hope that helped
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3 years ago
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NaClO3 &gt; NaCl + O2 <br> Balance
tiny-mole [99]

Answer:

Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.

6 0
3 years ago
if the percent by volume is 2% and the volume of solution is 250 mL what is the volume of solute and solution
Liula [17]

Answer:

Solute = 5 mL; solution = 250 mL  

Explanation:

The formula for percent by volume is

\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 2 % v/v,

\begin{array}{rcl}2 \, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\2 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{2 \times 250\text{ mL}}{100}\\\\ & = & \textbf{5 mL}\\\end{array}

If there is no change of volume on mixing,

Volume of solution = 250 mL

 -Volume of solute = <u>     </u><u>5</u><u>      </u>

 Volume of solvent = 245 mL

5 0
3 years ago
A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 l to 2.00 l at constant temperature. after the compressi
FinnZ [79.3K]
Answer: 16 atm  
Explanation: 
P1V1 = P2V2 
P2 = P1V1/V2 
=4 atm x 8.00 L/2.00L = 16 atm
5 0
3 years ago
Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con
arsen [322]

Answer:

a)  T_b=590.775k

b)  W_t=1.08*10^4J

d)  Q=3.778*10^4J

d)  \triangle V=4.058*10^4J

Explanation:

From the question we are told that:

Moles of N2 n=2.50

Atmospheric pressure P=100atm

Temperature t=20 \textdegree C

                      t = 20+273

                     t = 293k

Initial heat Q=1.36 * 10^4 J

a)

Generally the equation for change in temperature is mathematically given by

\triangle T=\frac{Q}{N*C_v}

  Where

  C_v=Heat\ Capacity \approx 20.76 J/mol/K

T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }

T_b-293k=297.775

T_b=590.775k

b)

Generally the equation for ideal gas is mathematically given by

 PV=nRT

For v double

 T_c=2*590.775k

 T_c=1181.55k

Therefore

PV=Wbc

Wbc=(2.20)(8.314)(1181_590.778)

Wbc=10805.7J

Total Work-done W_t

W_t=Wab+Wbc

W_t=0+1.08*10^4

W_t=1.08*10^4J

c)

Generally the equation for amount of heat added is mathematically given by

Q=nC_p\triangle T

Q=2.20*2907*(1181.55-590.775)\\

Q=3.778*10^4J

d)

Generally the equation for change in internal energy of the gas is mathematically given by

\triangle V=nC_v \triangle T

\triangle V=2.20*20.76*(1181.55-293)k

\triangle V=4.058*10^4J

3 0
2 years ago
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