c) potassium, K(s)
b) manganese, Mn(s)
d) boron, B(s)
Answer:
The molar mass of:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
Explanation:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.
Answer:
1. The coefficients are: 1, 3, 2
2. From the balanced equation, we obtained the following:
3 moles oxygen, O2 reacted.
2 moles of Hydrogen sulfide, H2S reacted.
2 moles of water were produced.
2 moles of sulphur dioxide, SO2 were produced.
Explanation:
1. Determination of the coefficients of the equation.
This is illustrated below:
P2O5(s) + H2O(l) <==> H3PO4(aq)
There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:
P2O5(s) + H2O(l) <==> 2H3PO4(aq)
There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:
P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)
Now the equation is balanced.
The coefficients are: 1, 3, 2.
2. We'll begin by writing the balanced equation for the reaction. This is given below:
2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)
From the balanced equation above,
3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.
In the balanced chemical equation provided, 3 moles of oxygen react with 2 moles of hydrogen sulfide to produce 2 moles of water and 2 moles of sulfur dioxide.
When the balanced chemical equation 2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g) is considered, we can deduce that 3 moles of oxygen and 2 moles of hydrogen sulfide react together in this reaction. The products of this chemical reaction are 2 moles of water and 2 moles of sulfur dioxide. Each of these quantities is directly inferred from the coefficients in front of each compound in the balanced chemical equation.
Answer:- 1500 calories
Solution:- mass of bear = 1.850 g
volume of water = 100.0 mL
Density of water is 1.00 g/moL. So, mass of water would be 100.0 g.
delta T for water = 15.0 degree C
specific heat capacity for water is 1 cal/(g* degree C)
q = m x c x delta T
where, q is the heat energy, m is mass, c is specific heat capacity and delta T is change in temperature.
for water, q = 100.0 x 1 x 15.0
q = 1500 calorie
heat gained by water = heat lost by bear
So, the 1.850 g bear has 1500 cal or 1.50 Cal.
(Where, 1 Cal = 1000 cal)
Explanation:
Acids are the species which furnish protons (hydrogen ions) when dissolved in the water.
Bases are the species which furnish hydroxide ions when dissolved in the water.
Oxoacid is the acid which contains with at least one hydrogen atom which is bonded to the oxygen atom in the molecule which can dissociate in the solution to give proton and the corresponding anion.
Oxoanion is the anion which is derived from oxoacid by the loss of hydrogen atom which is bounded to the oxygen.
A hydrate is the specie which contains water molecule or it's constituents in its solid structure.
Onoble gases
O halogens
O transitional metals
Group/Family 18 on the periodic table is called the noble gases.
Group/Family 18 on the periodic table is called the noble gases. The noble gases are a group of chemical elements that have full valence electron shells, which makes them stable and nonreactive. This group includes elements like helium, neon, argon, krypton, xenon, and radon.
A H2A and OH− A2− and OH−
B HA− and Li+ A2− and Li+
C HA− and OH− HA− and Li+
D H2A and Li+ HA− and OH−
The main components (more than half of the initial amount of H2A, besides H2O) at equivalence point 1 and 2 (EP1), (EP2) is HA⁻ and Li⁺ A²⁻ and Li⁺.
Titration of acid H₂A with LiOH solution.
At first equivalent points are:
H₂A + LiOH → HA⁻ + Li⁺ + H₂O
The Main component at (EP1) => HA⁻ and Li⁺
At second equivalence point are:
HA⁻ + LiOH → A²⁻ + Li⁺ + H₂O
Main component at (EP2) => A²⁻ and Li⁺
Therefore, the correct option is B which is HA− and Li+ A2− and Li+
Find more information about Titration curve here:
Answer:
B) HA⁻ and Li⁺ A²⁻ and Li⁺
Explanation:
Titration of acid H₂A with LiOH solution.
At first equivalent point
H₂A + LiOH → HA⁻ + Li⁺ + H₂O
Main component at (EP1) => HA⁻ and Li⁺
At second equivalence point
HA⁻ + LiOH → A²⁻ + Li⁺ + H₂O
Main component at (EP2) => A²⁻ and Li⁺
Therefore, the correct answer is B