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Contact [7]
3 years ago
11

On the periodic table, in which period is copper?

Chemistry
1 answer:
Firdavs [7]3 years ago
3 0

Answer:

At the top of Group 11 above silver and gold.

Period 4

Explanation:

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Describe transfer RNA job.<br> pls thank you
Scorpion4ik [409]

Answer:

yes

Explanation:

5 0
3 years ago
Using the periodic table as a reference, which elements could bond to form a compound most like NaCl?
pychu [463]

<u>step</u><u> </u><u>by</u><u> </u><u>step</u>

Na(sodium)=2.8.1

Cl (Chlorine)=2.8.7. sodium will give the chlorine the 1 valence electron to become stable ions.

<u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>

<u>p</u><u>o</u><u>t</u><u>t</u><u>a</u><u>s</u><u>i</u><u>u</u><u>m</u><u>. </u><u>(</u><u>2</u><u>.</u><u>8</u><u>.</u><u>8</u><u>.</u><u>1</u><u>)</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>F</u><u>l</u><u>o</u><u>u</u><u>r</u><u>i</u><u>n</u><u>e</u><u>(</u><u>2</u><u>.</u><u>7</u><u>)</u>

4 0
3 years ago
Consider the following hypothetical molecular collisions and predict which of the following will form potential products, given
Annette [7]

Answer:

The collision theory is defined as the rate of a reaction is proportional to the rate of reactant collisions.

Explanation:

The reacting species should collide with orientation that allows contract between the atoms that will become bonds together in the product.

The collision occurs with adequate energy  to permit mutual penetration of the reacting species. The two physical factors based on the orientation and energy of collision, the following reaction with carbon monoxide with oxygen is considered.

 2CO(g) + O2(g) → 2CO2 (g)

After collision between the carbon monoxide and oxygen the reaction is

 CO(g) + O2(g) → CO2 (g) + O(g)

Based on the theories of chemical reaction the molecules collide with sufficient amount of energy an activated complex is formed.

7 0
3 years ago
A 46.2 mL,0.568 M calcium nitrate solution is mixed with 80.5mL of 1.396M calcium nitrate solution.Calculate tge concentration o
Ne4ueva [31]

Answer:

1.09 M

Explanation:

Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

c=\frac{n}{V}

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

n_{Ca(NO_3)_2}=46.2 mL\cdot0.568 M+80.5 mL\cdot1.396 M=138.62 mmol

Now, the total volume of this solution can be found by adding the volume values of each component:

V_total=46.2 mL+80.5 mL=126.7 mL

Finally, dividing the moles found by the total volume will yield the final molarity:

c_{final}=\frac{138.62 mmol}{126.7 mL}= 1.09 M

6 0
3 years ago
4. How many grams of ammonium carbonate are needed to decompose in order to produce
Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

4 0
3 years ago
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