<h3>
Answer:</h3>
- Balanced Equation; 2Fe + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂
- Why balance?: To obey the law of conservation of mass
- Subscripts can not be changed, since they show the actual number of atoms of each element in a compound.
<h3>
Explanation:</h3>
- The balanced equation for the reaction between the iron metal and sulfuric acid to produces iron (III) sulfate and hydrogen gas is given by;
2Fe + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂
- Balancing chemical equations ensures that they obey the law of conservation of mass which requires the mass of reactants and the mass of products to be equal.
- Balancing is done by putting coefficients on reactants and products while not affecting the subscripts as subscripts show the actual number of atoms of an element in a compound.
Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.
The ruling separation is
d=1 / (470mm −1) = 2.1×10⁻³ mm
Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.
Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.
We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is, find the greatest integer value of m for which mλ<d.
since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3
that value is m=3.
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
Learn more about diffraction here : brainly.com/question/16749356
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I think the answer is 98 grams
Answer:
pH = 10.9
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:

The resulting concentrations are:
![[CO_3^{2-}]=\frac{0.1435mol}{0.25L}=0.574M \\](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B0.1435mol%7D%7B0.25L%7D%3D0.574M%20%5C%5C)
![[HCO_3^{-}]=\frac{0.0265mol}{0.25L}=0.106M](https://tex.z-dn.net/?f=%5BHCO_3%5E%7B-%7D%5D%3D%5Cfrac%7B0.0265mol%7D%7B0.25L%7D%3D0.106M)
Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:

Which makes sense since basic OH⁻ ions were added.
Regards!