Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
Answer:
The answer to your question is: letter D.
Explanation:
Noble gases are located in group VIIIA of the periodic table, this means that they have 8 eight electrons in their outermost shell.
Due to this characteristic, they are stable and do not react with other elements.
a. 1s22s22p4 The outermost shell of this electron configuration has 6 electrons, then this element has 6 electrons not 8. This configuration is of an element of the group VIA.
b. [Ne]2s22p2 The outermost shell of this element has 4 electrons, so this is not the configuration of a noble gas.
c. [Ar] 3s1 This element only has one electron in its outermost shell, so this is the electron configuration of an alkaline metal.
d. 1s22s22p6 This element has 8 electrons in its outermost shell, so this is the electron configuration of a noble gas.
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg
Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol
The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺
The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5
Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol
Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg
Answer:
Explanation:
We know that momentum is the product of mass and velocity so here
mass (m) = 1700 kg
velocity (v) = 13 m/s
So now
momentum = m * v
= 1700 * 13
= 22100 kg m/s
hope it helps :)
