Answer:
See explanation
Explanation:
The ionic radius of metal M decreases as the charge on the metal M increases. The ionic radius is generally defined as the distance between the nucleus and the outermost electron of the ion. Hence, ionic radius becomes much lesser as the magnitude of the positive charge increases.
It is obvious from the various formulae of metal chlorides in the question that the metal forms cations M^2+, M^3+ and M^4+ respectively. The order of decreasing ionic radius of the compounds is;
MX2 > MX3 > MX4
Answer: The final concentration of aluminum cation is 0.335 M.
Explanation:
Given: 
= 47.8 mL (1 mL = 0.001 L) = 0.0478 L
= 0.321 M, 
= 21.8 mL = 0.0218 L, 
= 0.366 M
As concentration of a substance is the moles of solute divided by volume of solution.
Hence, concentration of aluminum cation is calculated as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D)
Substitute the values into above formula as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%5C%5C%3D%20%5Cfrac%7B0.321%20M%20%5Ctimes%200.0478%20L%20%2B%200.366%20M%20%5Ctimes%200.0218%20L%7D%7B0.0478%20L%20%2B%200.0218%20L%7D%5C%5C%3D%20%5Cfrac%7B0.0153438%20%2B%200.0079788%7D%7B0.0696%7D%5C%5C%3D%200.335%20M)
Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.
Answer:
Explanation:
magnesium + nitrogen ⟶ Product
13 g 5 g
Mass of product = 13 g + 5 g = 18 g
The product contains 5 g of nitrogen
.
Answer:
D
Explanation:
The indivisibility of an atom was proved wrong an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions......
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The product of the nuclear reaction in which 31p is subjected to neutron capture followed by alpha emission is ²⁸Al.
Nuclear
reaction: ³¹P + n° → ²⁸Al + α (alpha particle).<span>
Alpha decay is radioactive decay in which an atomic
nucleus emits an alpha particle (helium nucleus) and transforms
into an atom with an atomic number that is reduced by
two and mass number that is reduced by four.</span>
