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ladessa [460]
2 years ago
13

Write the empirical formula of a list for binary ionic compounds that could be formed from the following ions

Chemistry
1 answer:
klasskru [66]2 years ago
7 0

Answer:

CaBr₂, CaO, FeBr₂, FeO

Explanation:

Per binary compound, there must be 1 cation (Ca²⁺, Fe²⁺) and 1 anion (Br⁻, O²⁻). The compound should have a charge of 0 (neutral). For this to occur, there sometimes needs to be more than one cation/anion present in the compound. Below I have included the math that displays why there are more ions necessary in some compounds.

CaBr₂

-----> Ca = +2 and Br = -1

-----> +2 + (-1) + (-1) = 0

CaO

-----> Ca = +2 and O = -2

-----> +2 + (-2) = 0

FeBr₂

-----> Fe = +2 and Br = -1

-----> +2 + (-1) + (-1) = 0

FeO

-----> Fe = +2 and O = -2

-----> +2 + (-2) = 0

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When you look out the window you can see constantly moving air.Air is cycled in order to reach equilibrium
aleksley [76]
False, although the air is moving in an equilibrium us normal humans can not see the air actually moving. So it is false.
6 0
3 years ago
A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon
svetlana [45]

Answer:

\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}

Explanation:

1. Miles travelled in an average month  

\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}

2. Using a gasoline powered vehicle  

(a) Moles of heptane used  

n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}

(b) Equation for combustion  

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O  

(c) Moles of CO₂ formed  

n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}

(d) Volume of CO₂ formed  

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.  

V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}

3. Using an electric vehicle  

(a) Theoretical energy used  

\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}

(b) Actual energy used  

The power station is only 85 % efficient.  

\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

The heat of combustion of methane is -802.3 kJ·mol⁻¹  

V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}

4. Comparison  

\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}

6 0
3 years ago
Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH3NH2) with 0.110 M HCl. The Kb of methylam
nikklg [1K]
The reaction between the reactants would be:

CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻

Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.

             CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I                0.11                       0             0
C               -x                          +x           +x
E            0.11 - x                     x             x

Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]

Since the given information is Kb, let's find Ka in terms of Kb.

Ka = Kw/Kb, where Kw = 10⁻¹⁴

So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]

Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>


4 0
3 years ago
Question 5
pantera1 [17]

Answer: The coefficient for the diatomic oxygen (O2) is 3.

Explanation:

To know the coefficient for the diatomic Oxygen, we need to balance the equation.

Fe + O2 ------->   Fe2O3

LHS of the equation; Fe =  1    , O2 = 1

RHS of the equation; Fe = 2 ,  O = 3

∴ Multiply 'Fe' on the LHS of the equation by 4 and O2 by 3

   Doing that will give the balance equation which is;

   4 Fe +  3 O2  --------> 2 Fe2O3

The coefficient for the diatomic oxygen (O2) as seen from the equation is 3.

7 0
3 years ago
Calculate the density of an object with a mass of 6.147 and a volume of 9.3 (include units)
Olin [163]
Here is the formula for density:
Density (D) = Mass (M) divided by Volume (V)

So you would do D = 6.147 divided by 9.3

As an as answer you would get: 0.6609677419g/cm^3

Additional information:

The formula for volume is:
V = M divided by D


The formula for Mass is:
M = D times V


I hope this helps :)
5 0
3 years ago
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