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Alex
2 years ago
12

PLEASE ANSWER I NEED THE ANSWER ASAP!!!!!!!!!The table shows the amount of radioactive element remaining in a sample over a peri

od of time. Part 1: what is the half life of the element? Explain how you determined this.Part 2: how long would it take 308 g of the sample to decay to 4.8125 grams? Show your work or explain your answer? * please explain the answer is easy to understand terms, thank you*
Chemistry
1 answer:
qaws [65]2 years ago
3 0

Answer:

pt1

The half-life of the element is 1,600 years.

pt2

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1,600 years) = 4.33 x 10⁻⁴ year⁻¹.

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 4.33 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 312.0 g).

[A] is the remaining concentration of the sample ([A] = 9.75 g).

∴ t = (1/k) ln([A₀]/[A]) = (1/4.33 x 10⁻⁴ year⁻¹) ln(312.0 g/9.75 g) = [8,000 year].

-Hops

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Based on the products obtained, rank the functional groups (acetamido, amino, and methoxy) in order of increasing ability to act
I am Lyosha [343]

Answer:

Amino >Methoxy > Acetamido

Explanation:

Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.

The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.

The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.

Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.

Therefore, decreasing order will be as follows:

Amino >Methoxy > Acetamido

7 0
2 years ago
Identify the factors that directly favor the unloading of oxygen from hemoglobin in the blood near metabolically active tissues.
AlexFokin [52]

Answer:

A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES

B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.

C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.

Explanation:

Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.

There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.

The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.

There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.

The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.

7 0
2 years ago
HELP! Ethanol has a density of 0.8 g/cm3. a. What is the mass of 225 cm3 of ethanol? b. What is the volume of 75.0 g of ethanol?
IRINA_888 [86]

Answer:

The correct answers are:

a) 180 g

b) 93.7 cm³

Explanation:

The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:

density= mass/volume

From the data provided in the problem:

density = 0.8 g/cm³

a) Given: volume= 225 cm³

mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g

b) Given: mass= 75.0 g

volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³

5 0
1 year ago
A molecule contains 23.24 g iodine (I),
gavmur [86]
Convert each amount of grams into moles:

I: 23.24g x 1 mol / 126.90g = 0.1831 mol I

C: 2.198 x 1 mol / 12.01g = 0.1830 mol C

N: 2.562 x 1 mol / 14.01g = 0.1829 mol N

Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1

Therefore the empirical formula is ICN
7 0
1 year ago
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
2 years ago
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