Determine the number of protons, electron and neutrons present in an atom of magnesium

If it requires 48.3 milliliters of 0.55 molar nitric acid to neutralize 15.0 milliliters of barium hydroxide, solve for the mola

Gemiola [76]

First you have to moles so multiply .0483L X .55M= .026565 Multiply moles by mole ratio which is 1/2, so the moles becomes .013283 now molarity=moles/volume; divide .013283/.015L=.885533M significant figures and you final answer is 0.89M

8 0

3 years ago

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If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma

Veseljchak [2.6K]

Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

4 0

3 years ago

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves

hjlf

This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

$2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

$3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

Explanation :

The given two chemical reactions are:

(1) $2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

(2) $3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:

(1)

(2) $6MnO_2(s)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)$

Now we are adding both the reactions, we get the overall chemical reaction.

$6MnCO_3(s)+3O_2(g)+6MnO_2(s)+8Al(s)\rightarrow 6MnO_2(s)+6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

The $MnO_2$ is common on both side, by cancelling it, we get:

The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

5 0

3 years ago

Which one of the following pair has the same number of ions?

zhuklara [117]

Hahahaaaa none of the above but IF <span>(c) is

1/2 mole of NaCl and 1/3 mole of MgCl2 instead,

then C is the right ans :)</span>

6 0

3 years ago

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Describe briefly how to obtain the radial probability of an electron

Kaylis [27]

Explanation:

The radial distribution function gives the probability density for an electron to be found anywhere on the surface of a sphere located a distance r from the proton. Since the area of a spherical surface is 4πr2, the radial distribution function is given by 4πr2R(r)∗R(r).

I

8 0

2 years ago