Answer:
I promise it is the 2nd one
Explanation:
The following diagram shows a stage of a cell during mitosis. <u>Two identical circles are joined in the middle</u> and they have the nucleolus inside the nucleus in the center of the circles.
True. These ions are of a 2- charge. Oxygen is an example. It will form a 2- charge if ionize.
Answer:
a) 0.115 g
Explanation:
The balanced reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
To heat 55g of water, the energy in form of heat necessary can be calculated by:
Q = mcΔT
where Q is the heat, m is the mass, c is the specific heat (for water, c = 4.18 J/gºC), and ΔT is the variation of the temperature, which is 25ºC, so:
Q = 55x4.18x25
Q = 5747.5 J = 5.7475 kJ
So, for the reaction, 1 mol of CH₄ releases 802.3 kJ, so to release 5.7475 kJ will be necessary:
1 mol ---------------- 802.3 kJ
x ---------------- 5.7475 kJ
By a simple direct three rule:
802.3x = 5.7475
x = 7.164x10⁻³mol
The molar mass of CH₄ is : 12 (of C) + 4x1 (of H) = 16 g/mol
The mass is equal to the number of moles multiplied by molar mass, the:
m = 7.164x10⁻³x16
m = 0.115 g
It will go over the amount it needs to.
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
----------------------------------
Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
