write a balanced chemical equation for the reaction of zncl2 with excess NaOH to produce Na2Zn(oh)4 sodium zincate. what mass of sodium zincate can be produced from 2.00 g of ZNCl2 with excess Naoh by this reaction?
Answers
∴
=
= 0.0147 mol
1 : 1
∴
Mass = Molar Mass * Mol
∴
= 179 g/mol * 0.0147 mol
= 2.63 g
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Answers
1. Temperature and pressure is constant.
2. Ideal gas is involved.
The variables involved in this problem is the moles of the gas and the volume. We find their relationship from the ideal gas law. PV=nRT. With all other concept besides n and V is constant, Volume is directly proportional to the amount of gas inside the balloon. Solving this is as simple as ratio and proportion.
25.5L/3.5 moles = =15.4L/X moles
X= 2.11 moles.
Answer:
61 grams
Explanation:
If the number of moles of CO₂ in the balloon is given to be 3.5 moles, the mass of CO₂ in the balloon will be
number of moles = mass ÷ molar mass
Hence, mass = number of moles × molar mass
Molar mass of CO₂ is 44g. This is 44 because the atomic mass of carbon is 12 while that of oxygen is 16. Thus, 12 + (16 × 2) = 44 g
mass= 3.5 moles × 44g
mass = 154g (which is the initial mass)
When the mass of CO₂ in the 25.5 liter balloon is 154g, the mass of CO₂ in the balloon when the volume of CO₂ in the balloon is 15.4 liter will be X
To get X,
25.5 L ⇒ 154g
15.4 L ⇒ X
cross-multiply, and
X = (15.4 × 154) ÷ 25.5
X = 93.00 grams (which is the final mass)
93.00 grams of CO₂ was left in the balloon, hence the mass of CO₂ that escaped will be: initial mass minus final mass
= 154g - 93g
= 61g
The mass of CO₂ that escaped is 61 grams
Answers
Answer:
A substance which burns in air and tends to produce heat and light is known as Combustible substances. Non-combustible substances are certain substances which are not combustible in the presence of air. Will not burn on being exposed to flame.
Answer:
A substance which burns in air and tends to produce heat and light is known as Combustible substances. Non-combustible substances are certain substances which are not combustible in the presence of air. Will not burn on being exposed to flame.
Explanation:
Answers
Answer:
4 moles of Li₃N will be produced in this reaction
Explanation:
The reaction is:
6Li + N₂ → 2Li₃N
If the nitrogen gas is the excess reactant, the limiting must be the lithium.
You always have to make calculations with the limiting reactant. You never use the excess reagent.
Ratio is 6:2.
The rule of three to solve this is:
6 moles of lithium can produce 2 moles of nitride
Therefore, 12 moles of Li must produce (12 . 2) / 6 = 4 moles of nitride
Answer:
4.0 moles of lithium nitride will be produced
Explanation:
Step 1: Data given
Number of moles lithium (Li) = 12.0 moles
Nitrogen gas (N2) is in excess.
Step 2: The balanced equation
6Li + N2 → 2Li3N
Step 3: Calculate moles of lithium nitride (Li3N)
For 6 moles Lithium we need 1 mol nitrogen gas to produce 2 moles lithium nitride
For 12.0 moles lithium we'll have 12.0/ 6 = 2.0 moles nitrogen gas to react to produce 12.0 / 3 = 4.0 moles lithium nitride
4.0 moles of lithium nitride will be produced
(2) Pb-206 (4) Co-60
Answers
Final answer:
I-131 (Iodine-131) is the radioisotope used for diagnosing thyroid disorders. It is selectively taken up by the thyroid gland and its emissions allow for imaging of the gland.
Explanation:
The radioisotope used for diagnosing thyroid disorders is I-131 (Iodine-131). This radioisotope is used in medical procedures due to its unique characteristics. When consumed, it is selectively taken up by the thyroid gland. I-131 emits beta particles and gamma radiation which allows for imaging of the thyroid gland. This in turn helps health professionals diagnose if the thyroid is functioning normally or if disorders such as hyperthyroidism or tumors are present.
Learn more about I-131 Radioisotope here:
#SPJ11
a) [A] >> [B]
b)Changing the temperature will have no affect on the equilibrium.
c)There is twice as much A as there is B in the reaction vessel.
d)There is more than 100 times more B then there is A in the reaction vessel.
e)There is twice as much B as there is A in the reaction vessel.
Answers