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Elina [12.6K]
3 years ago
13

Descuss how to find the oldest paper in a stack of papers

Chemistry
2 answers:
deff fn [24]3 years ago
8 0
It is by looking on the dates of the paper.

Hoped this helped.

~Bob Ross®
MariettaO [177]3 years ago
8 0
Either by looking at the dates or seeing which paper looks old.
You might be interested in
The equilibrium constant, Kc, for the following reaction is 3.61×10-4 at 426 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently lar
iogann1982 [59]

Answer:

the concentration of PCl5 in the equilibrium mixture = 296.20M

Explanation:

The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.

The detaied steps is as shown in the attached file.

8 0
3 years ago
(WILL GIVE BRAINLESS )Complete the sentence describing Newton's third law of motion: For every action in one direction, O b ос O
Liono4ka [1.6K]

Answer:

An equal force acts in the opposite direction.

Explanation:

3 0
3 years ago
Determine the empirical formula of the compound formed when 1.2g of magnesium reacts with 3.55g of chlorine.Take the molar mass
masha68 [24]

Explanation:

For Mg, (1.2 g Mg/24 g Mg) = 0.05 mol Mg.

For Cl, (3.55 g Cl/35.5 g Cl) = 0.1 mol Cl

So the ratio now is

Mg:Cl = 0.05 : 0.1 = 1:2

I got the 1:2 ratio by dividing both by the smallest number, which is 0.05 mol. Therefore, the empirical for formula of the substance is MgCl_2

5 0
3 years ago
When the vapor pressure of water is 0.106 mmHg, determine the reaction quotient for that above equilibrium.
TEA [102]

Answer:

Answer is explained in the explanation section below.

Explanation:

Note: This question is not complete and lacks necessary data to solve. However, I have found a similar question and I will be using its data to solve this question for the sake of understanding and concept.

Solution:

Equilibrium Reaction:

CaO(s) + H2O(g) -->Ca(OH)2(s)    

We need to find the reaction quotient for this question:

Q = \frac{1}{P_{H20} }

Here, only the pressure of the gaseous reactant will be used and here H20 is the only reactant which is gaseous.

And we are given that, vapor pressure of water is = 0.106 mmHg

So,

Now, we need to convert it into atm

so, 1atm = 760 mmHg

0.106 mmHg = 0.106/760 atm

0.106 mmHg = 1.394 x 10^{-4} atm

Plugging in the values in the equation, we get:

Q = \frac{1}{P_{H20} }

Q = \frac{1}{1.394 . 10^{-4} }

Q = 7173.60 atm^{-1}

8 0
3 years ago
H2SO4 + 2NaNO2 → 2HNO2 + Na2SO4
dem82 [27]
<h3>Answer:</h3>

23.459 g NaNO₂

<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>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄

[Given] 24.14714 g Na₂SO₄

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

[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂

Molar Mass of Na - 22.99 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol

Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 24.14714 \ g \ Na_2SO_4(\frac{1 \ mol \ Na_2SO_4}{142.05 \ g \ Na_2SO_4})(\frac{2 \ mol \ NaNO_2}{1 \ mol \ Na_2SO_4})(\frac{69.00 \ g \ NaNO_2}{1 \ mol \ NaNO_2})
  2. Multiply/Divide:                                                                                                \displaystyle 23.4587 \ g \ NaNO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We need 5 sig figs (instructed).</em>

23.4587 g NaNO₂ ≈ 23.459 g NaNO₂

8 0
3 years ago
Read 2 more answers
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