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Studentka2010 [4]
3 years ago
13

A chemical equation is shown below. KNO3 → KNO2 + O2 What are the coefficients that should be added to balance this equation? Us

e complete sentences to explain your answer. Explain how this chemical reaction demonstrates the conservation of mass. Question 21 (Essay Worth 8 points) (04.03 MC) Part 1: Name the type of chemical reaction that occurs when magnesium chloride solution (MgCl2) reacts with sodium carbonate solution (Na2CO3). Part 2: Explain why aluminum (Al) would react with copper chloride (CuCl2) but not with magnesium chloride (MgCl2). Question 22 (Essay Worth 8 points) (04.07 HC) The table shows the amount of radioactive element remaining in a sample over a period of time. Radioactive Decay Rate Amount of Radioactive Sample (grams) Time (years) 45.0 0 31.8 11 22.5 22 15.9 33 11.3 44 Part 1: What is the half-life of the element? Explain how you determined this. Part 2: How long would it take 308 g of the sample to decay to 4.8125 grams? Show your work or explain your answer.
Chemistry
1 answer:
uysha [10]3 years ago
7 0

20) Balanced chemical equation: 2KNO₃ → 2KNO₂ + O₂.

Potassium nitrate is decomposed on potassium nitrite and oxygen.

According to principle of mass conservation, number of atoms must be equal on both side of chemical reaction.  

There are two potassium atoms, two nitrogen atoms and six oxygen atoms on both side of balanced chemical reaction.  

Decomposition is reaction where one substance (in this example potassium nitrate) is broken down into two or more simpler substances (potassium nitrite and oxygen).  

21) The answer is:

Balanced chemical reaction: double recplacement reaction.

MgCl₂(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2NaCl(aq).

The magnesium (Mg) and the sodium (Na) replace each other making this a double replacement reaction.

Reactivity series is an empirical progression of a series of metals, arranged by their reactivity from highest to lowest (alkaline metals have highest reactivity and Noble metals lowest reactivity):  

K > Ba > Sr > Na > Ca > Mg > Be > Al > Mn > Zn > Cr> Fe > Cd > Co > Ni > Sn > Pb > H(in acids) > Cu > Hg > Ag.

Metal higher in the reactivity series will displace another and it will be oxidized.

Magnesium is higher it this series than aluminium and aluminium is higher than copper.

22) The answer is: half life is 22 years.

Half-life is the time required for a quantity (in this example mass of the element) to reduce to half its initial value and is independent of initial concentration.

Mass of the element is reduced by half in 22 years from 45 grams to 22.5 grams.

First calculate the radioactive decay rate constant λ:

λ = 0.693 ÷ t.

λ  = 0.693 ÷ 22 y.

λ = 0.0315 1/y.

ln(m/m₀) = -λ·t₁.

ln(4.8125/308) = -0.0315 1/y · t₁.

t₁ = 132.02 y.

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
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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
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
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