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3 years ago

The crystal structure of an ionic compound depends on the

Chemistry

1 answer:

Murljashka [212]3 years ago

4 0

Answer:

The ion's sizes and their radius ratio.

Explanation:

Ionic compounds take the form of three dimentional arrays by alternating anions and cations in order to minimize the potential energy of the system by maximizing the attractive force between opposite charges.

The resultant geometric structures are known as crystal lattices and their arrangements depend on the ion's sizes and, since they are bound though electrostatic attraction, their radius ratio.

I hope you find this information useful and interesting! Good luck!

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Exactly 10.0 L of air -25°C is heated to 100.0°C. What is the new volume if the pressure is kept constant?

zepelin [54]

Answer: V2= 15.0403226 Liters

Explanation:

Use V1/T1=V2/T2

Make sure you change the degrees Celsius to Kelvin. (Kelvin = degrees Celsius +273)

10.0L / 248 K = V2/ 373 K

Cross multiply V1 and T2 and divide by T1

(10.0 L)( 373K)/ 248 K = V2

V2= 15.0403226 Liters (Kelvin cancels out)

4 0

3 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

2 years ago

Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to

snow_lady [41]

Answer:

$\displaystyle \text{p} K_a \approx 3.856$

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:

$\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right] & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}$

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

$\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}$

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for pKₐ:

$\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}$

In conclusion, the pKₐ value of lactic acid is about 3.856.

5 0

1 year ago

How many moles of N2O5 are needed to produce 7.90 g of NO2?

Roman55 [17]

Answer:

0.085 moles of N₂O₅ are needed

Explanation:

Given data:

Mass of NO₂ produces = 7.90 g

Moles of N₂O₅ needed = ?

Solution:

2N₂O₅ → 4NO₂ + O₂

Number of moles of NO₂ produced :

Number of moles = mass/ molar mass

Number of moles = 7.90 g/ 46 g/mol

Number of moles = 0.17 mol

now we will compare the moles of NO₂ with N₂O₅.

NO₂ : N₂O₅

4 : 2

0.17 : 2/4×0.17 = 0.085 mol

Thus, 0.085 moles of N₂O₅ are needed.

4 0

2 years ago

how many grams of oxygen gas are needed to produce 10.0 grams of carbon dioxide according to the balanced equation of CH4

Alinara [238K]

Molar mass of :

O2 = 16 * 2 = 32 g/mol

CO2 = 12 + 16 * 2 = 44 g/mol

Balanced chemical equation :

1 CH4 + 2 O2 = 1 CO2 + 2 H2O
↓ ↓
2 moles 1 mole

2* 32 g O2 ----------> 1* 44 g CO2
x g O2 ------------> 10.0 g CO2

44 x = 2 * 32*10.0

44 x = 640

$x = \frac{640}{44}$

$x = 14.54 g$ of O2

7 0

3 years ago