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jeka94
2 years ago
10

Calcium-45 has a half-life of 162.7 days. If four half-lives have elapsed, how much time has passed?

Chemistry
2 answers:
vovikov84 [41]2 years ago
7 0
Half-life<span> is the time required for a quantity to reduce to half its initial value. </span><span>If four half-lives have elapsed for calcium-45, then it would be 4x162.7 = 650.8 days have passed. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
Len [333]2 years ago
6 0

Answer:

1.78 years

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Fourth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

There are four half lives. So, time has passed = 4*162.7 days = 650.8 days = 1.78 years

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In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OAT
dmitriy555 [2]

Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

\Delta G=\Delta H-T\Delta S

\Delta G = Gibbs free energy  

\Delta H = enthalpy change

\Delta S = entropy change  

T = temperature in Kelvin

When \Delta G = +ve, reaction is non spontaneous

\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

For the given reaction 1: glucose+Pi\rightarrow glucose-6-phosphate+H_2O \Delta G=+13.8kJ/mol

As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.

6 0
2 years ago
Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio
AlexFokin [52]

Answer:

The equilibrium concentration of CH₃OH is 0.28 M

Explanation:

For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)

The equilibrium constant (Keq) is given for the following expresion:

Keq= \frac{(CH3OH)}{(CO) x (H2)^{2}} =14.5

Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.

We have:

(CH3OH)= ?

(CO)= 0.15 M

(H2)= 0.36 M

So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:

14.5= \frac{(CH_{3}OH) }{(0.15 M) x (0.36 M) ^{2} }

14.5 x (0.15 M) x (0.36)^{2} = (CH₃OH)

0.2818 M = (CH₃OH)

6 0
2 years ago
You need to prepare 2.00 L of 0.100 M Na2CO3 solution. The best procedure is to weigh out: ___________
jeka94

Answer:

The answer to your question is letter C

Explanation:

Data

Volume = 2 L

Molarity = 0.100 M

Molecular weight Na₂CO₃ = (2 x 23) + (1 x 12) + (3 x 16)

                                           = 46 + 12 + 48

                                           = 106 g

Process

1.- Calculate the grams of Na₂CO₃ needed

                         106 g ----------------  1 mol

                           x      ----------------  0.1 moles

                           x = (0.1 x 106) / 1

                           x = 10.6 g

2.- Calculate the grams of Na₂CO₃ needed for 2 liters of solution

                           10.6 g -------------- 1 liter

                            x        --------------  2 liters

                           x = (10.6 x 2) / 1

                           x = 21.2 grams of Na₂CO₃                        

                     

4 0
2 years ago
Read 2 more answers
What are some non examples of biodiversity
zhenek [66]
A monocrop is a non example of biodiversity because it contains only one species, such as all corn, therefore there is very little biodiversity.
8 0
2 years ago
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3. A 0.500 g sample of nitrogen gas combines with 1.140 g of oxygen gas to form NO2. If the atomic mass of oxygen is 16.000, cal
DedPeter [7]

Answer;

= 18.24

Explanation;

The ratio of N and O in the formula NO2 IS 1:2

Mass of nitrogen gas is 0.500 g

Moles of nitrogen will be;

= 0.500/16 = 0.03125 moles

Therefore;

The moles of Oxygen from the ratio will be;

= 0.03125 × 2 = 0.0625 moles

But; 0.0625 moles is equal to 1.140 g of Oxygen

The atomic number (mass in 1 mole) will be;

= 1.140 /0.0625

= 18.24

Thus the atomic number of Oxygen from the data is 18.24

6 0
2 years ago
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