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Kaylis [27]
3 years ago
6

How many grams of O2 are required to produce 5 mol CO2

Chemistry
1 answer:
maxonik [38]3 years ago
7 0
I did the math i got 220 grams
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2. As NH4OH is added to an HCl solution, the pH of the solution
chubhunter [2.5K]

Answer:

c

Explanation:

Nh4OH + HCL ---> NH4Cl + H3O

so ph decreases as H3O increases

and OH also decreases

5 0
3 years ago
Read 2 more answers
202H8(9) + 7 O2(g) > 4CO2(g) + 6 H2O(9).
lilavasa [31]

since the concentration of Carbon Dioxide will increase, it would make Q > K, cause equilibrium to shift in the direction with less moles of gas to alleviate the extra pressure. In this case, the reaction will shift left because there are fewer moles of gas present.

4 0
3 years ago
What is the mass of 3.77mol of K3N?
AleksAgata [21]
<h3>Answer:</h3>

495 g K₃N

<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>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

3.77 mol K₃N

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

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

  1. Set up:                       \displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})
  2. Multiply/Divide:         \displaystyle 495.039 \ g \ K_3N

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

5 0
3 years ago
how do the properties of compounds compare to the properties of the elements from which they are composed
Crank
<span>The elements that form the compound don't compare because it doesn't matter about the elements when they are combined. The elements that formed it together is completely different than the outcome compound. </span>
6 0
3 years ago
Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at
telo118 [61]

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

H_2(g)+I_2(g)\rightarrow 2HI(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound            Pressure at equilibrium

H_2                                   61.8 atm

I_2                                    46.5 atm

HI                                  52.3 atm

Calculate the value of the equilibrium constant K_p for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant K_p for this reaction is, 0.952

Explanation :

The given chemical reaction :

H_2(g)+I_2(g)\rightarrow 2HI(g)

The expression of K_p for above reaction follows:

K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}

We are given:

P_{H_2}=61.8atm

P_{I_2}=46.5atm

P_{HI}=52.3atm

Putting values in above equation, we get:

K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952

Therefore, the value of equilibrium constant K_p for this reaction is, 0.952

7 0
3 years ago
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