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3 years ago

How many protons are in a radium-226 nuclide? A. 88 B. 138 C. 226 D. 314

Chemistry

2 answers:

True [87]3 years ago

5 0

Answer: Option (A) is the correct answer.

Explanation:

Atomic number is defined as the sum of total number of protons present in an element.

Whereas mass number is defined as the sum of total number of protons and neutrons present in an atom.

Also, when an atom is neutral in nature then its number of protons are equal to the number of electrons.

For example, atomic number of radium is 88 and mass number is 226.

Hence, number of neutrons can be calculated as follows.

Mass number = no. of protons + no. of neutrons

226 = 88 + no. of neutrons

No. of neutrons = 226 - 88

= 138

Thus, we can conclude that there are 88 protons present in radium-226 nuclide.

Maksim231197 [3]3 years ago

3 0

²²⁶₈₈Ra
........................
protons : 88
electrons : 88
neutrons : 226-88 = 138
..................
A) 88

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3 years ago

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Suppose of potassium iodide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of potassium c

swat32

Answer:

0.0983 M

Explanation:

First, we need to find the formulas of the reactants. Potassium forms the ion K⁺, and iodide is the ion I⁻, thus potassium iodide is KI. Silver forms the ion Ag⁺, and nitrate is the ion NO₃⁻, thus silver nitrate is AgNO₃. In the reaction, the cations will be replaced:

KI(aq) + AgNO₃(aq) → KNO₃(aq) + AgI(s)

AgI is an insoluble salt, so it will precipitate, and all nitrates are soluble, thus KNO₃ will be in the ionic form: K⁺ and NO₃⁻. 1 mol of KNO₃ = 1 mol of K⁺.

The molar mass of KI is 166 g/mol, thus the number of moles that is added is:

nKI = mass/molar mass

nKI = 5.71/166 = 0.0344 mol

And the number of moles of AgNO₃ is given as 64mM = 0.064 mol. Because the stoichiometry is 1:1, AgNO₃ is in excess, thus, all the KI will react and form 0.0344 mol of KNO₃. So, nK⁺ = 0.0344 mol. The molarity is the number of moles divided by the volume (350 mL = 0.350 L):

0.0344/0.350 = 0.0983 M

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3 years ago

The following equation shows the equilibrium in an aqueous solution of ammonia: NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH3(aq)+H2O(l)⇌NH

yarga [219]

Answer:

c) H2O and OH−

Explanation:

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Thus, for the reaction,

$NH_3+H_2O\rightleftharpoons Nh_4^++OH^-$