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Maru [420]
4 years ago
12

Please Helpppppp!!!!!!!

Chemistry
1 answer:
stich3 [128]4 years ago
8 0

Answer:

2. D

3. B

4. B

Explanation:

These are what make sense to me.

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How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -&gt;<br> 4A1 + 302
Evgesh-ka [11]
<h3>Answer:</h3>

\displaystyle 40 \ mol \ Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 40 \ mol \ Al

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0
3 years ago
Determine the root-mean square speed of CO2 molecules that have an average kinetic energy of 4.2 x10-21 J per molecule.
strojnjashka [21]
<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2 

If you substitute into this formula, you will get out the root-mean-square speed. 

If energy is Joules, the mass should be in kg, and the speed will be in m/s. 

1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg. 

If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule. 

4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg 

So, if E = 1/2 mv^2 

</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68 
Take the square root of that, and you get the answer 339 m/s.
8 0
4 years ago
How many particles are in 12.47 grams of NaCl?
jeka94
If we are talking about moles then the answer to that is 0.22
5 0
3 years ago
Which characteristic of living organisms is described by the following definition? "The ability
saw5 [17]
3 I don’t know but it may be movement
3 0
3 years ago
Enthalpy of <br><br> CH4(g) + 2NO2(g) -&gt; N2(g) + CO2(g) + 2H2O(l)
stira [4]

Answer:

-177.9 kJ.

Explanation:

Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.

5 0
3 years ago
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