<u>Answer:</u> The products of the given chemical equation are 
<u>Explanation:</u>
Protonation equation is defined as the equation in which protons get added in the substance.
The chemical equation for the protonation of carbonate ion in the presence of water follows:

By Stoichiometry of the reaction:
1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion
Hence, the products of the given chemical equation are 
Answer:they have the same number of electrons in the valence shell
Explanation:
Energy were released from the walnut, q = 1,673.6 J
<h3>Equation :</h3>
To find the energy using formula,
q = mcΔt
where,
q is charge
m is mass
c is specific heat of water
Δt is change in temperature
So, given
t₁ = 50°C
t₂ = 60°C
m = 40g
c = 4.184 J/g
Now putting the values known,
We get,
q = mc(t₂ - t₁)
q = 40g x 4.184 J/g x (60 - 50)
q = 167.36 J x 10
q = 1,673.6 J
<h3>What is heat energy?</h3>
Heat is the thermal energy that is transferred when two systems with different surface temperatures come into contact. Heat is denoted by the letters q or Q and is measured in Joules.
To know more about specific heat :
brainly.com/question/11297584
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Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample
Answer:
5 000 000 (5 million atoms)
Explanation:
Let us assume that a vanadium atom has a spherical shape.
diameter of a sphere = 2 x radius of the sphere
Thus,
Radius of a vanadium atom = 130 pm
= 130 x
m
The diameter of a vanadium atom = 2 x radius
= 2 x 130 x
= 260 x
m
Given a distance of 1.30 mm = 1.30 x
m,
The number of vanadium atoms required to span the distance = 
= 5000000
Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.