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NemiM [27]
2 years ago
12

20 POINTS I will give brainliest to whoever answers correctly

Chemistry
1 answer:
Romashka [77]2 years ago
6 0

All organisms need four types of organic molecules: nucleic acids, proteins, carbohydrates and lipids; life cannot exist if any of these molecules are missing.

You might be interested in
How many grams will be produced??
Akimi4 [234]

Explanation:

b is correct. 30.6 g H2O is produced.

3 0
3 years ago
Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the form
musickatia [10]

Answer:

This is the balanced equation:

Pb(NO₃)₂ (aq) + 2NaI (aq) → 2NaNO₃ (aq)  +  PbI₂ (s) ↓    

Explanation:

This are the reactants:

PbNO₃

NaI

Iodide can react to Pb²⁺ to make a solid compound.

4 0
3 years ago
Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
The reaction of glucose, c6h12o6, with oxygen produces carbon dioxide and water. what is the sum of the coefficients in the bala
Misha Larkins [42]

The balanced equation for the given reaction:

C₆H₁₂O₆ (glucose) + 6O₂→ 6CO₂ + 6H₂O

So in the balanced equation the coefficients before glucose, oxygen, water and carbon dioxide are 1, 6, 6 and 6 respectively.

Therefore, the sum of the coefficients in the balanced equation

= 1 + 6 + 6 + 6

= 19

The correct answer is 19.

8 0
3 years ago
With initial consent for a warrantless search, police spent three days searching a
xz_007 [3.2K]

Michigan v. Tyler was the Supreme court ruling that they violated when

warrantless search was done.

<h3>Michigan v. Tyler case</h3>

The Court of Appeals of the State of Michigan affirmed the conviction for the

exception of a warrantless search only in the case of an arson.

In this case, there was no arson involved which is against the rule of law as

various items (stolen money, illegal weapons, drugs) were only searched for.

This thereby violates the U.S. Supreme Court ruling between Michigan v.

Tyler.

Read more about Michigan v. Tyler case here brainly.com/question/1622038

3 0
2 years ago
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