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Sveta_85 [38]
3 years ago
10

HELP!!

Chemistry
1 answer:
alexandr402 [8]3 years ago
4 0

Answer:

21: True

22: False

23: True

24: False

25: False

26: True

27: True

28: False

30: True

Explanation: Hope this is helpful :)

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For each of the scenarios, identify the order with respect to the reactant, A. A⟶products The half‑life of A increases as the in
nirvana33 [79]

Answer:

Answers are in the explanation.

Explanation:

  • The half‑life of A increases as the initial concentration of A decreases. order: <em>2. </em>In the half-life of second-order reactions, the half-life is inversely proportional to initial concentration.
  • A three‑fold increase in the initial concentration of A leads to a nine‑fold increase in the initial rate. order: <em>2. </em>The rate law of second-order is: rate = k[A]²
  • A three‑fold increase in the initial concentration of A leads to a 1.73‑fold increase in the initial rate. order: <em>1/2. </em>The rate law for this reaction is: rate = k √[A]
  • The time required for [A] to decrease from [A]₀ to [A]₀/2 is equal to the time required for [A] to decrease from [A]₀/2 to [A]₀/4. order: <em>1. </em>The concentration-time equation for first-order reaction is: ln[A] = ln[A]₀ - kt. That means the [A] decreasing logarithmically.
  • The rate of decrease of [A] is a constant. order: <em>0. </em>The rate law is: rate = k -<em>where k is a constant-</em>
4 0
3 years ago
How much of a sample remains after five half-lives have occurred?
timama [110]

Answer:

1/32 of the original sample

Explanation:

We have to use the formula

N/No = (1/2)^t/t1/2

N= amount of radioactive sample left after n number of half lives

No= original amount of radioactive sample present

t= time taken for the amount of radioactive samples to reduce to N

t1/2= half-life of the radioactive sample

We have been told that t= five half lives. This implies that t= 5(t1/2)

N/No = (1/2)^5(t1/2)/t1/2

Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;

N/No= (1/2)^5

N/No = 1/32

Hence the fraction left is 1/32 of the original sample.

3 0
3 years ago
Make a diagram showing how communication satellites could be used to relay
alexdok [17]

Answer:

Explanation:

Satellite acts as a relay station for earth sources. A transmission station sends the information to the satellite, which in turn retransmits to the receiving stations. The original signal being transmitted from the earth station to the satellite is called uplink of 6 GHz The retransmitted signal from the satellite to the receiving station is called downlink of 4 GH frequency. The transponder is the transmitter-receiver combination in a transponder.

6 0
3 years ago
Which of the following would most likely be reduced when combined with
PSYCHO15rus [73]

Answer:

This is based on most likely so it's A.Fe

5 0
3 years ago
The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The
dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

3 0
3 years ago
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