The best description of Ernest Rutherford's experiment is letter C. The positively charged particles were fired through a gold foil.
Both noticed that different atoms gives different colours of light when they are exposed to flame.
S orbital can hold 2
P can hold 6
D can hold 10
F can hold 14
1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.
How to calculate number of atoms?
The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
However, the number of moles of oxygen in glycine can be calculated using the following expression:
Molar mass of C₂H5O2N = 75.07g/mol
Mass of oxygen in glycine = 32g/mol
Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine
Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles
Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms
Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.
Learn more about number of atoms at: brainly.com/question/8834373
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Answer:
The volume on the tank is 6, 20 L
Explanation:
We use the formula PV=nRT. We convert the units of pressure in kPa into atm and temperature in Celsius into Kelvin:
0°C=273K
101,325kPa---1 atm
275kPa --------x=(275kPax 1 atm)/101,325kPa= 2,71 atm
PV=nRT --> V=nRT/P
V= 0,750 mol x 0,082 l atm /K mol x 273 K/ 2, 71 atm= <em>6, 20 L</em>
