Their are 18 covalent bonds in a molecule of cyclohexane
Answer:
The answer to your question is 0.269 g of Pb
Explanation:
Data
Lead solution = 0.000013 M
Volume = 100 L
mass = 0.269 g
atomic mass Pb = 207.2 g
Chemical reaction
2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)
Process
1.- Calculate the mass of Pb in solution
Formula
Molarity = 
Solve for number of moles
Number of moles = Volume x Molarity
Substitution
Number of moles = 100 x 0.000013
Number of moles = 0.0013
2.- Calculate the mass of Pb formed.
207.2 g of Pb ----------------- 1 mol
x g ----------------- 0.0013 moles
x = (0.0013 x 207.2) / 1
x = 0.269 g of Pb
Answer:
827 mL
Explanation:
To answer this question we use the <em>definition of Molarity</em>:
Molarity = mol / L
[Cl⁻] = mol Cl⁻ / L
Now we calculate the moles of Cl⁻ present in 42.0 g of MgCl₂⋅6H₂O:
Molar mass of MgCl₂⋅6H₂O = 24.3 + 2*35.45 + 6*18 = 203.2 g/mol
moles of Cl⁻ = 42.0 g MgCl₂⋅6H₂O ÷ 203.2 g/mol *
= 0.4134 mol Cl⁻
Finally we use the definition of molarity to calculate the volume:
0.500 M = 0.4134 mol Cl⁻ / xL
xL = 0.827 L = 827 mL
Answer:
43.2 g of table salt occupies 20.0 cm2of space- Appropriate for calculating density
150 g of iron density of iron 79.0 g/cm3- Appropriate for calculating volume
21.0 mL of methanol density of methanol 0.79g/mL - Appropriate for calculating mass
3.00 g of white pine, density of white pine 0.40 g/cm3- Appropriate for calculating volume
5.00 cm,3 of table salt weighs 10.8g-Appropriate for calculating density
Explanation:
Density is defined as mass per unit volume. Hence in science;
Density= mass/volume
When mass and volume are both given, the information can be used to determine the density of a given substance.
When volume and density are both given, the information can be used to determine the mass of a given substance.
When the mass and density of a substance are both given, the information can be used to determine the volume of the substance.
Hence the answers given above.
Answer:
Cytoplasm
Explanation:
Cytoplasm is a jellylike substance within cells.