Answer:
FeCl₃ + <u>3</u>NaOH → Fe(OH)₃ + <u>3</u>NaCl
Explanation:
You are given a skeleton equation, or unbalanced equation.
FeCl₃ + NaOH → Fe(OH)₃ + NaCl
<u>Count the number of atoms on each side</u>. If there is 1 atom, the subscript is not written. If the subscript to the right of a bracket, it belongs to each element in the bracket.
Reactants → Products
Fe 1 Fe 1
Cl 3 Cl 1 Not balanced
Na 1 Na 1
O 1 O 3 Not balanced
H 1 H 3 Not balanced
<u>The number of each element must be the same in the reactants and products.</u> We can see that Cl (chlorine), O (oxygen) and H (hydrogen) have different numbers.
<u>Write in coefficients in front of formulas</u> (separated by + sign). Coefficients multiply each subscript (or "1" if not written) in the formula. <u>Recount the atoms until you have equal reactant and product atom numbers.</u>
I see that I need more chlorine in my products. Multiply the formula that has "Cl" by 3.
FeCl₃ + NaOH → Fe(OH)₃ + 3NaCl
Reactants → Products
Fe 1 Fe 1
Cl 3 Cl 3 Cl is balanced now.
Na 1 Na 3 Not balanced
O 1 O 3 Not balanced
H 1 H 3 Not balanced
Fix the "Na" now. Multiply the formula with "Na" on the left by 3.
FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
Reactants → Products
Fe 1 Fe 1
Cl 3 Cl 3
Na 3 Na 3
O 3 O 3
H 3 H 3
Each side has the same number of each element. Therefore, this is balanced now.
Balanced formula:
FeCl₃ + <u>3</u>NaOH → Fe(OH)₃ + <u>3</u>NaCl
Answer:
a. Hydrocarbons have low boiling points compared to compounds of similar molar mass.
b. Hydrocarbons are hydrophobic.
d. Hydrocarbons are insoluble in water.
Explanation:
As we know that the hydrocarbons is a mix of carbon and hydrogen. In this the availability of the electronegative atom is not there that shows there is no bonding of the hydrogen plus it is dissolved. Also, the hydrocarbons is considered to be a non-polar but as compared to the water, water is a polar
In addition to this, the strong bond is no existed that shows the lower boiling points
Therefore option A, B and D are right
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Empirical formula is the simplest way the molecular formula can be wrote so here 7 goes into all of these so it would be CH2O
Answer:
the stronger light 5.5 m apart from the total illumination
Explanation:
From the problem's statement , the following equation can be deducted:
I= k/r²
where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality
denoting 1 as the stronger light and 2 as the weaker light
I₁= k/r₁²
I₂= k/r₂²
dividing both equations
I₂/I₁ = r₁²/r₂²=(r₁/r₂)²
solving for r₁
r₁ = r₂ * √(I₂/I₁)
since we are on the line between the two light sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus
r₂ = r₁ + d
then
r₁ = (r₁ + d)* √(I₂/I₁)
r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)
r₁*(1-√(I₂/I₁)) = d*√(I₂/I₁)
r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁)) =
r₁ = d/[√(I₁/I₂)-1)]
since the stronger light is 9 times more intense than the weaker
I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3
then since d=11 m
r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m
r₁ = 5.5 m
therefore the stronger light 5.5 m apart from the total illumination
