Answer:
38.7%
41.3%
20%
Explanation:
The percentage composition helps to know the what percent of the total mass of a compound is made up of each of the constituent elements or groups.
To solve this problem:
Compound:
Ca₃(PO₄)₂
Formula mass = 3(40) + 2[31 + 4(16)]
= 120 + 2(95)
= 120 + 190
= 310
%C = x 100 = 38.7%
%P = x 100 = 41.3%
%O = x 200 = 20%
pa help po,,,
Answer:
Endangered. The meaning of endangered is when a species declines in number- this might be because they are being eaten by another animal which has a higher popupayion, or perhaps because they are being hunted by humans for their meat or fur. Neither reason might be that their food or habitat has been destroyed b humans to create land for farming or housing. one example of name endangered species is Polar Bears: they like in the Antarctic where there is no land, simply ice. Due to humans causing climate change, the world is warming up and the ice is melting. This means polar bears are dying out because it would be too tiring to swim to find ice. For the same reason, other animals are dying in the Antarctic meaning that Poplar Bears have less to eat (being at the top of the food chain -no animal eats polar bears).
Answer:
Please help?????
Explanation:
4.92*10-3=4.92/10^3=
4.92/1000=0.00492
Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
To determine the moles of KIO_3 titrated, use the balanced equation 2 KIO_3 + 5 Na_2S_2O_3 + 6 HCl → 3 I_2 + 6 NaCl + 6 NaClO + 3 H_2O. Therefore, 0.001551 mol of KIO_3 were titrated.
To determine the moles of KIO3 titrated, we need to use the balanced equation for the reaction:
2 KIO3 + 5 Na2S2O3 + 6 HCl → 3 I2 + 6 NaCl + 6 NaClO + 3 H2O
From the equation, we can see that 2 moles of KIO3 react with 5 moles of Na2S2O3. Therefore, the moles of KIO3 titrated can be calculated using the following proportion:
(0.0100 M KIO3 / 1 L) * (15.51 mL / 1000 mL) * (2 mol KIO3 / 5 mol Na2S2O3) = 0.001551 mol KIO3
#SPJ12
general formula
Answer:
Explanation:
Hello!
In this case, since the heat of vaporization is related with the energy required by a substance to undergo the phase transition from liquid to gas, we can compute such amount of energy as shown below:
In such a way, since the enthalpy of vaporization is given as well as the mass, we compute the energy as shown below:
Best regards!
Answer:
34 g
Explanation:
Let's consider the following balanced equation.
N₂ + 3 H₂ → 2 NH₃
The theoretical mass ratio of N₂ to H₂ is 28g N₂ : 6g H₂ = 4.6g N₂ : 1g H₂.
The experimental mass ratio of N₂ to H₂ is 100g N₂ : 6g H₂ = 16.6g N₂ : 1g H₂.
As we can see, hydrogen is the limiting reactant.
According to the task, we 6 g of H₂ react completely, 34 g of ammonia are produced.