Be sure to answer all parts. Rubidium and bromine atoms are depicted at right. Answer the following questions. (a) What is the c
harge on the rubidium ion? What is the charge on the bromide ion? (b) To which noble gas is the rubidium ion related? To which noble gas is the bromide ion related? h5sil8102 (c) Which pair below best represents the relative ionic sizes?
1 answer:
The complete question is found on the image attached.
Answer:
a) Rb= +1, Br= -1
b)Bromide and rubidium ions are related to krypton
c) C
Explanation:
Rubidium is found in group 1 with a charge of +1. Bromine is found in group 17 with a charge of -1. Both species have the same configuration as krypton. The ionic sizes of the ions are a very similar to each other Rb+ is 166 while Be is 167.
This RbBr is an ionic compound.
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Answer:
The molecular formula =
Explanation:
% of C = 11.79
Molar mass of C = 12.0107 g/mol
<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>
% of Cl = 69.57
Molar mass of Cl = 35.453 g/mol
<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>
Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,
% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64
Molar mass of F = 18.998 g/mol
<u>% moles of F = 18.64 / 18.998 = 0.9812</u>
Taking the simplest ratio for C, Cl and F as:
0.9816 : 1.9623 : 0.9812
= 1 : 2 : 1
The empirical formula is =
Also, Given that:
Pressure = 21.3 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 21.3 / 760 = 0.02803 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 458 mL = 0.458 L (1 mL = 0.001 L)
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 0.00052445 moles
Given that :
Amount = 0.107 g
Molar mass = ?
The formula for the calculation of moles is shown below:
Thus,
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol
Molar mass = 204.0233 g/mol
So,
Molecular mass = n × Empirical mass
204.0233 = n × 101.9147
⇒ n = 2
<u>The molecular formula = </u>
If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.
<h3>What is Stoichiometry ?</h3>Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.
<h3>What is Balanced Chemical Equation ?</h3>The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now we have to write the balanced equation
Mg + 2HNO₃ → Mg(NO₃)₂ + H₂
According to Stoichiometry
= 3.3 g H₂
Thus from the above conclusion we can say that If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.
Learn more about the Stoichiometry here: brainly.com/question/16060223
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