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kotykmax [81]
3 years ago
10

(a) ne (b) li (c) al (d) cl (e) ca 1. which of the elements above is most commonly found as a negatively charged ion?

Chemistry
1 answer:
LiRa [457]3 years ago
3 0
Cl because it gains electron and become negatively charged
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Heather has to do a titration with the acid-base reaction. She would need a complete buret assembly that contains, a buret, a buret clamp and a ringstand. She would also need an erlenmeyer flask and of course, a volumetric pipet with its pipet bulb.

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_________________ enter the bloodstream through the thin ______________________.
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cool

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Give the nuclear symbol (isotope symbol) for the isotope of bromine that contains 44 neutrons per atom. nuclear symbol:_________
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In the nuclear symbol of an element, the superscript represents the mass number. The mass number is the sum of the number of protons and the number of neutrons present in the atom. The subscript is the atomic number which is the number of protons present in the atom or the number of electrons in the neutral atom.

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How many grams of zinc metal will react completely with 7.8 liters of 1.6 M HCl? Show all of the work needed to solve this probl
Andru [333]

<u>Answer:</u>

For 1: The correct answer is 407.97 grams.

For 2: The correct answer is 76.72 grams.

<u>Explanation:</u>

  • <u>For 1:</u>

We are given the molarity of HCl, to find the moles of HCl, we use the formula:

Molarity=\frac{\text{Moles}{\text{Volume}}

We are given:

Molarity = 1.6 M

Volume = 7.8 L

Putting values in above equation, we get:

1.6mol/L=\frac{\text{Moles of HCl}}{7.8L}\\\\\text{Moles of HCl}=12.48mol

For the given reaction:

Zn(s)+2HCl(aq.)\rightarrow ZnCl_2(aq.)+H_2(g)

By Stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of zinc metal.

So, 12.48 moles of HCl react with = \frac{1}{2}\times 12.48=6.24mol of Zinc metal.

To calculate the mass of zinc metal, we use the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}    .....(1)

Molar mass of Zinc metal = 65.38 g/mol

6.24=\frac{\text{Mass of zinc metal}}{65.38g/mol}\\\\\text{Mass of zinc metal}=407.97g

  • <u>For 2:</u>

We are given that 298 grams of 83.1 % by mass of iron (II) nitrate solution are present.

So, mass of Iron (II) nitrate solution will be = \frac{83.1}{100}\times 298=247.638g

Molar mass Iron (II) nitrate = 180 g/mol

Putting values in equation 1, we get:

\text{Moles of }Fe(NO_3)_2=\frac{247.638g}{180g/mol}=1.37mol

For the following reaction:

2Al(s)+3Fe(NO_3)_2(aq.)\rightarrow 3Fe(s)+2Al(NO_3)_3(aq.)

By Stoichiometry of the reaction:

3 moles of Fe(NO_3)_2 produces 3 moles of iron metal.

So, 1.37 moles of Fe(NO_3)_2 will produce = \frac{3}{3}\times 1.37=1.37mol of iron metal.

To calculate the mass of zinc metal, we equation 1:

1.37=\frac{\text{Mass of iron metal}}{56g/mol}\\\\\text{Mass of iron metal}=76.72g

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