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Evgen [1.6K]
3 years ago
11

Please?????????? ????

Chemistry
1 answer:
natka813 [3]3 years ago
3 0
The causes of mass extinction
As there might be marks on fossils showing the suffering that happened to it!
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What is the half-life (in seconds) of a zero-order reaction which has an initial reactant concentration of 0.884 M with a k valu
pantera1 [17]
<span>Answer: 8.15s
</span><span />

<span>Explanation:
</span><span />

<span>1) A first order reaction is that whose rate is proportional to the concenration of the reactant:
</span><span />

<span>r = k [N]
</span><span />

<span>r = - d[N]/dt =
</span><span />

<span>=> -d[N]/dt = k [N]
</span><span />

<span>2) When you integrate you get:
</span><span />

<span>N - No = - kt
</span>
<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />

<span>=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k
</span><span />

<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
</span>
<span /><span /><span>
t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s</span>
6 0
3 years ago
Read 2 more answers
Which of the following is NOT a fate for pyruvate?
son4ous [18]

Answer: The conversion to malate  

Explanation:

Pyruvate is the process which produced in glycoysis which has multiple fates and it can give rises to acetyl co-enzyme and undergo the aerobic oxidation in the critic acid cycle. It can be used to produces glucose but it never produced the malate. In prokaryotes it can be processes in the anaerobic respiration to produced the ethanol, as end product.

6 0
3 years ago
How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
3 years ago
Pl help me, i need a good grade on this.
Blizzard [7]
1. Option A. Temperature and Salinity
2. Option D. (I think)
3. Option D.
Hope your test goes well!
8 0
3 years ago
Which section of the reaction represents the reactants
Diano4ka-milaya [45]
A should be the products and D should be the reactants. So D should be the answer.
8 0
3 years ago
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