Answer:
creat and trough are the parts of transverse wave
Hey there!:
1 mole of C6H12O6 ------------------ 6 moles of C
3.0 moles of C6H12O6 ------------- ??
3.0 *6 / 1 =>
18.0 moles of C
Hope that helps!
Answer:
1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
Explanation:
Given data:
mass of cadmium = 6.35 g
Number of atoms of aluminum as 6.35 g cadmium contain = ?
Solution:
Number of moles of cadmium = 6.35 g/ 112.4 g/mol
Number of moles of cadmium = 0.06 mol
Number of atoms of cadmium:
1 mole = 6.022×10²³ atoms of cadmium
0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol
0.36×10²³ atoms of cadmium
Number of atoms of Al:
Number of atoms of Al = 0.36×10²³ atoms
1 mole = 6.022×10²³ atoms
0.36×10²³ atoms × 1 mol /6.022×10²³ atoms
0.06 moles
Mass of aluminum:
Number of moles = mass/molar mass
0.06 mol = m/ 27 g/mol
m = 0.06 mol ×27 g/mol
m = 1.62 g
Thus, 1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
The density of urine decreases when a diabetic secretes too much sugar.
When there is excess sugar in the blood, the kidneys release extra water.
The result is a <em>more dilute</em> and therefore <em>less dense urine</em>.
The urine from a non-diabetic person usually has a density of 1.010 g/mL to 1.025 g/mL.
The urine from a diabetic person often has a density less than 1.008 g/mL.
Millimeters are an SI unit of length that =0.0001 m
