Answer:
Mass of oxygen in glucose = 29.3g
Explanation:
Mass of glucose given is 55grams.
We are to find the mass of oxygen in this compound.
In the compound we have 6 atoms of oxygen.
Solution
To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.
Atomic mass of C = 12g
H = 1g
O = 16g
Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180
Mass of O in glucose = 
x 55
= 
x 55
= 0.53 x 55
Mass of oxygen in glucose = 29.3g
Answer:
(NH₄)₃PO₄ (l) + Al(NO₃)₃(l) -----------→ AlPO₄(l) + 3NH₄NO₃(l)
Explanation:
Data Give:
Reaction between ammonium phosphate solution and solution of aluminum nitrate
- Write a balanced chemical equation
Details:
To write a balanced chemical equation we have to know formula units of compounds or molecules
Formula units
ammonium phosphate : (NH₄)₃PO₄
aluminum nitrate: Al(NO₃)₃
ammonium nitrate: NH₄NO₃
Now to write a chemical equation
- we have to write the chemical formulas or formula unit of each compound
- write the reactant on left side of the arrow
- write the product on the right side of the arrow
- put a plus sign in 2 reactants and products on each side of the arrow
- balance the equation by putting coefficient with compound formula
- write the phase symbols on the right corner of the compound formula in brackets
So the Reaction will be
(NH₄)₃PO₄ + Al(NO₃)₃ -----------→ AlPO₄ + NH₄NO₃
Now balance the Chemical equation
(NH₄)₃PO₄ + Al(NO₃)₃ -----------→ AlPO₄ + 3NH₄NO₃
Now write the phase Symbols
(NH₄)₃PO₄ (l) + Al(NO₃)₃(l) -----------→ AlPO₄(l) + 3NH₄NO₃(l)
all compounds in the reaction are in liquid form and soluble in water
*** Note:
There is no aluminum nitrite in chemicals formulas
Also ammonium nitrite can not be used in pure isolated form due to its highly instability
It may just be my internet but the picture is just white. maybe try reposting this..?
<h2>
<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u></h2>
<em>The female part is the pistil. The pistil usually is located in the center of the flower and is made up of three parts: the stigma, style, and ovary. The stigma is the sticky knob at the top of the pistil. It is attached to the long, tube</em><em> </em><em>like structure called the style</em><em>.</em>
<h3>
<em><u>H</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>I</u></em><em><u>t</u></em><em><u> </u></em><em><u>W</u></em><em><u>i</u></em><em><u>l</u></em><em><u>l</u></em><em><u> </u></em><em><u>H</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u> </u></em><em><u>Y</u></em><em><u>o</u></em><em><u>u</u></em><em><u> </u></em><em><u>!</u></em></h3>
The equilibrium constant is 1.3 considering the reaction as written in the question.
<h3>Equilibrium in chemical reactions</h3>
In a chemical reaction, the equilibrium constant is calculated based on the equilibrium concentration of each specie. The equation of this reaction is;
A (g) + 2B (g) ⇌ 3C (g).
The initial concentration of each specie is;
- A - 9.22 M
- B - 10.11 M
- C - 27.83 M
The equilibrium concentration of B is 18.32 M
We now have to set up the ICE table as follows;
A (g) + 2B (g) ⇌ 3C (g)
I 9.22 10.11 27.83
C -x -x +x
E 9.22 - x 10.11 - x 27.83 + x
The equilibrium concentration of B is 18.32 M hence;
10.11 - x = 18.32
x = 10.11 - 18.32 = -8.21
Hence;
Equilibrium concentration of A = 9.22 - (-8.21) = 17.43
Equilibrium concentration of C = 27.83 + (-8.21) = 19.62
Equilibrium constant K = [19.62]^3/[17.43] [18.32]^2
K = 1.3
Learn more about equilibrium constant: brainly.com/question/17960050
