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Karolina [17]
3 years ago
5

HELP. NO FAKE ANSWERS. I WILL REPORT. I AM CONFUSED AND NEED HELP. FILL IN THE NOT FILLED BOXES POR FAVOR.

Chemistry
1 answer:
SCORPION-xisa [38]3 years ago
7 0
Question #1
Potasium hydroxide (known)
 volume used is 25 ml 
Molarity (concentration) = 0.150 M
Moles of KOH used 
           0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4) 
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l) 
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
      0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid)  concentration
    0.001875 moles × 1000/15  
        = 0.125 M

Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
      0.25 × 30/1000
        =  0.0075 moles 
Sodium Hydroxide (base)
Volume used 20 ml 
Concentration (unknown)
The equation for the reaction is 
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr   is 1 : 1
Therefore, moles of NaOH used;
                 = 0.0075 moles
NaOH concentration will be 
       = 0.0075 moles × 1000/20
       = 0.375 M

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Balance this reaction:<br>C6H10Cl4+Cl2=​
Valentin [98]

Answer:

C6H10 + 2Cl2 → C6H10Cl4

Explanation:

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Why is petrified wood considered a fossil?
earnstyle [38]

Answer:

Petrified wood was part of a plant that has been changed into rocks.

Explanation:

Yeet

5 0
3 years ago
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A 0.5 mol sample of N2 is in a 6L container at 2 atm. what is the temperature of the gas in K
Andreas93 [3]

Answer:

300 K

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Diatomic Elements
  • Moles

<u>Gas Laws</u>

Ideal Gas Law: PV = nRT

  • <em>P</em> is pressure
  • <em>V</em> is volume
  • <em>n</em> is moles
  • <em>R</em> is gas constant
  • <em>T</em> is temperature

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>n</em> = 0.5 mol N₂

[Given] <em>V</em> = 6 L

[Given] <em>P</em> = 2 atm

[Given] <em>R</em> = 0.0821 L · atm · mol⁻¹ · K⁻¹

[Solve] <em>T</em>

<em />

<u>Step 2: Solve for </u><em><u>T</u></em>

  1. Substitute in variables [Ideal Gas Law]:                                                          (2 atm)(6 L) = (0.5 mol)(0.0821 L · atm · mol⁻¹ · K⁻¹)T
  2. Multiply [Cancel out units]:                                                                               12 atm · L = (0.04105 L · atm · K⁻¹)T
  3. Isolate <em>T</em> [Cancel out units]:                                                                             292.326 K = T
  4. Rewrite:                                                                                                             T = 292.326 K

<u>Step 3: Check</u>

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

292.326 K ≈ 300 K

8 0
3 years ago
If a particular ore contains 58.5% calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosph
alina1380 [7]
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Atomic weight:
Ca = 40  ; P = 31 ; O = 16

Ca3  = 40 * 3 = 120
P = 31
O4 = 16 * 4 = 64
(PO4)2 = (31 + 64) * 2 = 95 * 2 = 190

Ca3(PO4)2 = 120 + 190 = 310 g/mol

31/310 = 10% P in the calcium phospate

1.00kg * 1000 g/kg = 1000 g of phosphorus.

1000 g / 10% = 10,000 g of calcium phosphate

10,000 g / 58.5% = 17,094 grams of ore.

The minimum mass of the ore should be 17,094 grams or 17.094 kg.
5 0
3 years ago
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100 cm = 1m
shutvik [7]

Answer:

This is a conversion factor

3 0
2 years ago
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