I think they are primary producers, as they produce biomass from inorganic compounds (CO2)
Answer:
A forest or woodland area
Explanation:
Forests or woodland areas have lots of plants and grass like things that wuld burn easily
Answer:
C
Explanation:
'Ordered Arrangement' basically means that it is a solid at room temperature. Room temperature is approximately 15-20C so we are looking for melting and boiling points that are above room temperature so it hasn't/can't melt or boil at room temperature and would therefore be solid. Option C is the only one where both points are temperatures above room temperature therefore option C is the only one where the substance would be in an 'ordered arrangement' at room temperature.
Hope this helped!
Answer:
53.5g of NH4Cl
Explanation:
First, we need to obtain the number of mole of NH4Cl. This is illustrated below:
Volume = 0.5L
Molarity = 2M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 2 x 0.5
Mole = 1mole
Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:
Molar Mass of NH4Cl = 53.5g/mol
Number of mole = 1
Mass =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass = 1 x 53.5
Mass = 53.5g
Therefore, 53.5g of NH4Cl is contained in the solution.
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 %
