Answer:
A; C6H12O6 + 6O2
Explanation:
You can easily differentiate between the reactants and the products, as the reactants are on the left, while the products are on the right.
Answer:
Explanation:
3
Explanation:
The reaction expression is given as:
Al(OH)₃ + HNO₃ → H₂O + Al(NO₃)₃
To solve this problem, let us assign coefficient a,b,c and d to each specie;
aAl(OH)₃ + bHNO₃ → cH₂O + dAl(NO₃)₃
Conserving Al : a = d
O: 3a + 3b = c + 9d
H: 3a + b = 2c
N: b = 3d
let a = 1 , d = 1, b = 3 , c = 3
Multiply through by 3;
a = 1, b = 3, c = 3 and d = 1
Al(OH)₃ + 3HNO₃ → 3H₂O + Al(NO₃)₃
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
6H₂O + 6CO₂ + energy → C₆H₁₂O₆ + 6O₂
Explanation:
The given reaction represent the formation of glucose so it is photosynthesis reaction.
Photosynthesis:
It is the process in which in the presence of sun light and chlorophyll by using carbon dioxide and water plants produce the oxygen and glucose.
Carbon dioxide + water + energy → glucose + oxygen
water is supplied through the roots, carbon dioxide collected through stomata and sun light is capture by chloroplast.
Chemical equation:
6H₂O + 6CO₂ + energy → C₆H₁₂O₆ + 6O₂
it is known from balanced chemical equation that 6 moles of carbon dioxide react with the six moles of water and created one mole of glucose and six mole of oxygen.
<span>The transferred electron from lithium to fluorine provides each atom with a full outer energy level.</span>