Jill and Susan violated safety procedures by not properly listening and/or reading over the instructions to know all the materials, steps, and equipment they need for the lab. Hope this helps!
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 half-life of the compound is, 145 years.
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time passed by the sample = 60.0 min
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 25 = 75 g
Now put all the given values in above equation, we get
Now we have to calculate the half-life of the compound.
Therefore, the half-life of the compound is, 145 years.
Answer:
Enthalpy of formation of KF = -555 kJ/mol
Enthalpy of formation of CsF = -539kJ/mol
Explanation:
Explanations are provided in the attachment below
Answer:
sodium/potassium/rubidium/caesium/francium
Explanation:
all are group I elements, so they all have similar properties
