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nikdorinn [45]
3 years ago
10

What does it takes to prove hypothesis false

Chemistry
1 answer:
olganol [36]3 years ago
3 0

Answer:A hypothesis or model is called falsifiable if it is possible to conceive of an experimental observation that disproves the idea in question.

Explanation:That is, one of the possible outcomes of the designed experiment must be an answer, that if obtained, would disprove the hypothesis.

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10. What could the pH of a solution be if a cyanidin indicator turned blue? What is used to determine the endpoint of a titratio
Leviafan [203]

Answer:

The cyanidin indicator turns blue within a pH range of 5 - 7. The pH of the solution could be 5, 6 or 7.

An indicator is used to determine the endpoint of a titration.

Explanation:

Cyanidin indicator changes colour with each change in pH. In acidic solutions (pH < 7) cyanidin indicator will turn red, through to purple and blue, while in basic solutions (pH > 7), cyanidin indicator will change colour from aquamarine through to green and yellow. The cyanidin indicator turns blue within a pH range of 5 - 7.

Titration is a technique used in analytical chemistry to determine the unknown concentration of a solution. A solution of known concentration is added from a burette to the solution of unknown concentration until the reaction between the two solutions is complete. This known as the endpoint of the experiment. The endpoint of a titration is determined using an indicator which is added to reaction mixture. A colour charge is produced by the indicator at the endpoint of the reaction.

Note: An indicator is a dye of weak organic acids or bases which changes colour with changes in the pH of a solution. Some common indicators are methyl orange, methyl red, phenolphthalein, etc. These indicators are used to monitor the changes in the pH of solutions during a reaction.

7 0
3 years ago
A sample of liquid heptane (C7H16) weighing 11.5 g is reacted with 1.3 mol of oxygen gas. The heptane is burned completely (hept
Anna11 [10]

Answer:

a) 0.525 mol

b) 0.525 mol

c) 0.236 mol

Explanation:

The combustion reactions (partial and total) will be:

C₇H₁₆ + (15/2)O₂ → 7CO + 8H₂O

C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

---------------------------------------------------

2C₇H₁₆ + (37/2)O₂ → 7CO + 7CO₂ + 16H₂O

It means that the reaction will form 50% of each gas.

a) 0.525 mol of CO

b) 0.525 mol of CO₂

c) The molar mass of heptane is: 7*12 g/mol of C + 16*1 g/mol of H = 100 g/mol

So, the number of moles is the mass divided by the molar mass:

n = 11.5/100 = 0.115 mol

For the stoichiometry:

2 mol of C₇H₁₆ -------------- (37/2) mol of O₂

0.115 mol of C₇H₁₆ --------- x

By a simple direct three rule:

2x = 2.1275

x = 1.064 mol of O₂

Which is the moles of oxygen that reacts, so are leftover:

1.3 - 1.064 = 0.236 mol of O₂

5 0
4 years ago
What moving faster a golf ball or bowling ball
anygoal [31]
This all depends on lots of independent variables such as force used to roll the ball, wind resistance, etc. With no variables included, they would both move at the same velocity.
8 0
3 years ago
Read 2 more answers
Do all structural formulas for each amino acid have an amino group?
mina [271]
Amino acids are the monomers that make up proteins. Each amino acid has the same fundamental structure , which consists of a central carbon atom, also known as the alpha (α) carbon, bonded to an amino group (NH2), a carboxyl group (COOH), and to a hydrogen atom.
8 0
4 years ago
1.How many mL of 0.523 M HBr are needed to dissolve 8.60 g of CaCO3?
prohojiy [21]

Answer:

The answer to your question is:

Explanation:

1.-

HBr = 0.523 M   V = ?

CaCO3 = 8.6 g

                   2HBr(aq) + CaCO₃(s)     ⇒   CaBr₂(aq) + H₂O(l) + CO₂(g)

MW CaCO₃ = 40 + 12 + 48 = 100 g

MW HBr = 80 + 1 = 81 g

Molarity = moles / volume

                          100 g of CaCO₃ ----------------  1 mol

                            8.6 g                 ----------------   x

                            x = (8.6 x 1) / 100

                            x = 0.086 moles

                  2 moles of HBr ----------------- 1 mol of CaCO₃

                 x                         -----------------  0.086 moles

                 x = (0.086 x 2) / 1 = 0.172 moles of HBr

Volume = moles / molarity

Volume = 0.172/ 0.523 = 0.323 l or 323 ml of HBr

2.-

V = ? ml   NaOH 0.487 M

V = 101 ml of 0.628 M MnSO₄

                 MnSO₄(aq)  +  2NaOH(aq)  ⇒    Mn(OH)₂(s) + Na₂SO₄(aq)

MW MnSO₄ = 55 g

MW NaOH = NaOH = 40 g

Moles = Molarity x volume

Moles = (0.628) x (0.101)

Moles = 0.065 moles of MnSO₄

               1 mol of MnSO₄  ------------------ 2 moles of NaOH

               0.065                 -----------------   x

               x = (0.065x 2) / 1

              x = 0.131 moles of NaOH

Volume = moles / molarity

Volume = 0.131 / 0.487

Volume = 0.268 l or 268 ml of NaOH

4 0
3 years ago
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