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Olin [163]
3 years ago
13

What is the easiest way to find the charge of an element using the periodic table?!?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0
<span>On the Periodic Table metals are found on the left of the table and will be positive, while non-metals are found on the right of the table and will be negative. </span>
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What mass of copper is consumed in the reaction with hydrochloric acid if 10.0 g of gas is collected?
brilliants [131]

Answer:

317 g

Explanation:

Cu + 2HCl --> CuCl2 +H2

1      :     2           1 : 1

1 mole of Cu  = 63.5 g

1 mole of  H2 = 2g

1 mole  Cu produces = 1 mole of H2

63.5 g of Cu produces = 2 g of H2

So

10 g of H2 will be produced from =  (63.5/2)*10 = 317 g of Copper

4 0
3 years ago
The name given to the main group elements in group 2 is _______.
navik [9.2K]

Answer:

Alkaline Earth Metals

3 0
3 years ago
Doing it again please answer for me appreciate it thank you
Mandarinka [93]

Answer:

1. A

Explanation:

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4 0
3 years ago
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Gather data: On the CONTROLS pane, set the Reactant concentration to 2.0 mol/L, the Surface area to Maximum, and the Temperature
Vikentia [17]

Answer:

Explanation:

We can use the Arrhenius equation to relate the activation energy and the rate constant, k, of a given reaction:

k=Ae−Ea/RT

In this equation, R is the ideal gas constant, which has a value 8.314 J/mol/K, T is temperature on the Kelvin scale, Ea is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.

Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term,

e−Ea/RT, is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

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3 years ago
The solution of this problem with explaining the solution?
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I need a closer pic
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