The information that should be added to the chart in order to find out who ran a greater distance is THE UNITS USED TO MEASURE DISTANCE EACH DAY.
In science, the use of units if very important and each record made must be accompany with relevant units that are appropriate for the measurement taken. If units are not used along with data, it will be very difficult to quantify the values that are recorded. Scientists all over the world use SI units in their works in order to ensure uniformity.
The answer is potassium magnate
For 7A(17) :
Electronic configuration 
So, there are 5 unpaired electrons present in group 7A(17).
<h3>
What are Unpaired Electrons?</h3>
- An unpaired electron is an electron that doesn't form part of an electron pair when it occupies an atom's orbital in chemistry.
- Each of an atom's three atomic orbitals, designated by the quantum numbers n, l, and m, has the capacity to hold a pair of two electrons with opposing spins.
- Unpaired electrons are extremely uncommon in chemistry because an object carrying an unpaired electron is typically quite reactive. This is because the production of electron pairs, whether in the form of a chemical bond or as a lone pair, is frequently energetically advantageous.
- They play a crucial role in describing reaction pathways even though they normally only appear momentarily during a reaction on a thing called a radical in organic chemistry.
To learn more about unpaired electrons with the given link
brainly.com/question/14356000
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Answer: In simplest case mass of reactants is same as mass of products.
Without thinking this question deeper, mass of ZnCl2 would be 49, but..
Explanation: Reaction should be Zn + 2 HCl ⇒ ZnCl2 + H2
Amount of zinc is 5 g / 65,38 g/mol = 0,076476 mol and amount
of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.
Althought HCl is needed 0.152 moles, zinc is an limiting reactant.
So it is possible to produce only 0.076476 mol Hydrogen and its mass
is 0.154 g. Mass of ZnCl2 would be 0.076476 mol · (65.38 + 2·35.45) =
10.42 g
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
