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2 years ago

What are 5 of the critical events in evolution that occurred in the Paleozoic era?!!

Chemistry

1 answer:

soldi70 [24.7K]2 years ago

5 0

Answer:

Cambrian, Ordovician, Silurian, Devonian, Carboniferous, and Permian

Explanation:

These all occurred in the paleozoic Era I hope you do great good luck <3

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Which formula equation shows a reversible reaction?

pishuonlain [190]

Answer:2NaF is the correct one. It’s a simple combination and can be be split with relative ease

Explanation:

3 0

3 years ago

Read 2 more answers

The following data was collected when a reaction was performed experimentally in the laboratory.

REY [17]

Answer:

The answer to your question is 3 moles of AlCl₃

Explanation:

Process

1.- Write and balance the equation

Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃

2.- Determine the limiting reactant

Theoretical proportion 1 mol Al(NO₃)₃ : 3 moles of NaCl

Experimental proportion 4 moles Al(NO₃)₃ : 9 moles NaCl

From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃ increases four times.

3.- Determine the amount of AlCl₃ using proportions

3 moles of NaCl --------------- 1 mol of AlCl₃

9 moles of NaCl ---------------- x

x = (9 x 1) / 3

x = 9 /3

x = 3 moles

5 0

2 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Which of the following are quantized? A) canned soup from the grocery store b) the weight of jelly beans c) elevation of a perso

alisha [4.7K]

Both b) the weight of jelly beans and c) elevation of a person on a ramp

6 0

3 years ago

The neutrons within the nucleus of an atom have what charge?

vivado [14]

Answer:

the answer is D.

Explanation:

your welcome :)

4 0

2 years ago