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Kruka [31]
3 years ago
10

One method used commercially to peel potatoes is to soak them in a solution of NaOH for a short time and then remove them from t

he NaOH. The peel will have separated from the potato and so it can then easily be sprayed off to yield beautifully peeled potatoes! The NaOH solution however, must be analyzed periodically to make sure its concentration is adequate for the reaction In one such analysis, 45.7 mL of 0.500 M H2SO4 is required to react completely with a 20.0 mL sample of the NaOH solution. The products of the reaction are Na2SO4 and water. What is the concentration of the starting NaOH solution?a. 1.14 M b. 2.28 M c. 0.438 M d. 0.500 M e. 0.571M
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

b. 2.28 M

Explanation:

The reaction of neutralization of NaOH with H2SO4 is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

<em>Where 2 moles of NaOH react per mole of H2SO4</em>

<em />

To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:

<em>Moles H2SO4:</em>

45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4

<em>Moles NaOH:</em>

0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH

<em>Molarity NaOH:</em>

0.0457moles NaOH / 0.020L =

2.28M

Right option:

<h3>b. 2.28 M</h3>
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<h3>Answer:</h3>

64 g O₂

<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
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   36 g H₂O

[Solve]   x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \  H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Divide/Multiply [Cancel Units]:                                                                       \displaystyle 63.929 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

8 0
3 years ago
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An igneous rock has a coarse texture and is dark in color. How else can this rock be accurately described?
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Answer:

C,D.

Explanation:

Extrusive, or volcanic, igneous rock is produced when magma exits and cools above (or very near) the Earth's surface. These are the rocks that form at erupting volcanoes and oozing fissures.

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Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)
klio [65]

Answer:

Kp=3.07x10^6

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}

N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2

Consequently, the equilibrium constant is computed as:

Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6

Best regards.

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Answer:

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Consider the following system at equilibrium:
alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0
3 years ago
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