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Citrus2011 [14]
3 years ago
14

The diagram shows an animal cell.

Chemistry
1 answer:
LenKa [72]3 years ago
8 0

Answer:

X  

Explanation:

Assume your diagram is like the one below.

X represents a mitochondrion.

That's where the Tricarboxylic Acid Cycle converts a single glucose molecule into six molecules of CO₂.

W is wrong. It represents a vacuole, which can store both nutrients and waste products for later elimination.

Y is wrong. It represents the nucleolus, which plays a critical role in the synthesis of ribosomes.  

Z is wrong. It represents the cytoplasm, which is where cell processes like glycolysis and protein synthesis take place.

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Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h
Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

k=\frac{0.693}{t_1_/_2}

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

k=\frac{0.693}{1.5 hrs}

The expression for the rate constant is :

k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0
3 years ago
What are metalloids?
goldenfox [79]

A metalloid is a type of chemical element which has a preponderance of properties in between, or that are a mixture of, those of metals and nonmetals.

3 0
3 years ago
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What are bacteria? Chapter 7 lesson 2 AYU
Maru [420]

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Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significa
elena-s [515]

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

       Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I                                0                    0

C                              +S                +2S

E                                S                  2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

8 0
3 years ago
Potassium permanganate, KMnO, and glycerin, C3H5(OH)3, react explosively according to the
Finger [1]

The volume of CO2 at STP =124.298 L

<h3>Further explanation</h3>

Given

Reaction

4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O

701,52 g of KMnO4

Required

volume of CO2 at STP

Solution

mol KMnO4 (MW=158,034 g/mol) :

mol = mass : MW

mol = 701.52 : 158.034

mol = 4.439

mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549

At STP 1 mol = 22.4 L, so for 5.549 moles :

=5.549 x 22.4

=124.298 L

4 0
3 years ago
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