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OleMash [197]
3 years ago
12

Atoms bond because they become more ____ when they have ____ outer ______.

Chemistry
1 answer:
atroni [7]3 years ago
5 0
Atoms bing because the become more stable when they have eight outer electrons
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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

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I need help with this plz help
schepotkina [342]

Answer:

\rm ^{103}_{\phantom{1}40}Zr, zirconium-103.

Explanation:

In a nuclear reaction, both the mass number and atomic number will conserve.

Let ^{A}_{Z}\mathrm{X} represent the unknown particle.

The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle ^{A}_{Z}\mathrm{X}, A is the mass number while Z while Z is the atomic number.

Sum of mass numbers on the left-hand side of the equation:

\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240.

Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:

\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137.

Mass number conserves. As a result,

A + 137 = 240.

Solve this equation for A:

A = 103.

Among the five choices, the only particle with a mass number of 103 is \rm ^{103}_{\phantom{1}40}Zr. Make sure that atomic number also conserves.

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3 years ago
Atomic mass protons column electrons atomic number neutrons row 11 12 23
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A) Mass Number B) Protons and Electrons C) (I think group but I'm not 100% sure on this...) D) 12 (To work out the number of neutrons you do MASS number - ATOMIC number aka the number of protons so in this case you'd do 23-11=12) Hope this helps
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