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andreyandreev [35.5K]
2 years ago
7

For an atom of chlorine-35, what number would you put as the SUPERSCRIPT in the atomic notation?

Chemistry
1 answer:
kobusy [5.1K]2 years ago
6 0
<h2>Answer:</h2><h3><em>C</em><em>I</em><em>.</em><em>S</em><em>t</em><em>r</em><em>i</em><em>c</em><em>t</em><em>l</em><em>y</em><em> </em><em>s</em><em>p</em><em>e</em><em>a</em><em>k</em><em>i</em><em>n</em><em>g</em><em>,</em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>u</em><em>b</em><em>s</em><em>c</em><em>r</em><em>i</em><em>p</em><em>t</em><em> </em><em>i</em><em>s</em><em> </em><em>u</em><em>n</em><em>n</em><em>e</em><em>c</em><em>e</em><em>s</em><em>s</em><em>a</em><em>r</em><em>y</em><em> </em><em>,</em><em>s</em><em>i</em><em>n</em><em>c</em><em>e</em><em> </em><em>a</em><em>l</em><em>l</em><em> </em><em>a</em><em>t</em><em>o</em><em>m</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em>c</em><em>h</em><em>o</em><em>l</em><em>o</em><em>r</em><em>i</em><em>n</em><em>e</em><em> </em><em>h</em><em>a</em><em>v</em><em>e</em><em> </em><em>1</em><em>7</em><em> </em><em>p</em><em>r</em><em>o</em><em>t</em><em>o</em><em>n</em><em>s</em><em>.</em><em>H</em><em>e</em><em>n</em><em>c</em><em>e</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>i</em><em>s</em><em>o</em><em>t</em><em>o</em><em>p</em><em>e</em><em> </em><em>s</em><em>y</em><em>m</em><em>b</em><em>o</em><em>l</em><em>s</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>u</em><em>s</em><em>u</em><em>a</em><em>l</em><em>l</em><em>y</em><em> </em><em>w</em><em>r</em><em>i</em><em>t</em><em>t</em><em>e</em><em>n</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em>o</em><em>u</em><em>t</em><em>h</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>u</em><em>b</em><em>s</em><em>c</em><em>r</em><em>i</em><em>p</em><em>t</em><em>.</em><em>³</em><em>⁵</em><em>C</em><em>I</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>³</em><em>⁷</em><em>C</em><em>I</em></h3>

<h2><em><u>E</u></em><em><u>X</u></em><em><u>P</u></em><em><u>L</u></em><em><u>A</u></em><em><u>N</u></em><em><u>A</u></em><em><u>T</u></em><em><u>I</u></em><em><u>O</u></em><em><u>N</u></em><em><u>:</u></em></h2><h2><em>I</em><em> </em><em>H</em><em>P</em><em>E</em><em> </em><em>I</em><em>T</em><em> </em><em>H</em><em>E</em><em>L</em><em>P</em><em> </em><em>T</em><em>H</em><em>A</em><em>N</em><em>K</em><em> </em><em>Y</em><em>O</em><em>U</em><em>!</em></h2>

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A bowling ball rests on the floor. The bowling ball is given a push.
lyudmila [28]

Answer:

A. It will move.

8 0
2 years ago
How many grams of aluminum will be deposited by 0.1F? (Al=27) a.0.3g b. 0.9. c. 9.0g. d. 2.7g<br>​
andrezito [222]

Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.

<h3>What is electrolysis?</h3>

Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.

The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.

Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium

0.1 F will discharge = 0.1/3 × 27 g of aluminium

mass of aluminium deposited = 0.9 g of aluminium.

Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.

Learn more about electrolysis at: brainly.com/question/26050361

5 0
2 years ago
Calculate the vapor pressure of a solution made by dissolving 11.1 g Ca(OH)2 in 1 102 g of water at 25 °C. Vapor pressure of pur
Levart [38]

Answer:

23.15 mmHg.

Explanation:

To solve this question you need to understand that part of Raoult's law in your Chemistry textbook. So, let us delve right into the solution of the question.

We are given parameters such as the mass of Ca(OH)2 to be = 11.1 grams, mass of solvent = 102 grams, the Vapor pressure of pure water= 23.76 mm Hg, temperature = 25°C and vapour pressure of the solution= ??.

The molar mass of Ca(OH)2= 74 g/mol.

The first thing to do is to find the number of moles of Ca(OH)2 and that of water from the formula below;

Mass/ molar mass = Number of moles.

===> Number of moles, n= 11.1/ 74.

Number of moles = 0.15 moles Ca(OH)2.

===> Number of moles, n= 102/ 18.

Number of moles, n= 5.67 moles of water.

Next, we add the two moles together to find the solvent mile fraction since vapour pressure is proportional to mole fraction.

Then;

0.15 + 5.67 = 5.82.

Therefore, 5.67/ 5.82= 0.97.

Hence the vapor pressure of a solution = 0.97 × 23.76.

vapor pressure of a solution = 23.15 mmHg.

4 0
3 years ago
Read 2 more answers
In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
2 years ago
Name the two possible products in the precipitation reaction of copper (II) chloride and sodium phosphate. Use the charges on th
satela [25.4K]

Answer:

General equation for a double-displacement reaction:  

AB + CD --> AC + BD

• sodium chloride – NaCl copper sulfate – CuSO₄  

NaCl + CuSO₄ --> Na₂SO₄ + CuCl₂

The products formed are sodium sulfate and copper (II) chloride.

Copper (II) chloride forms a blue colored solution.

• sodium hydroxide – NaOH copper sulfate – CuSO₄  

NaOH + CuSO₄ --> Na₂SO₄ + Cu(OH)₂

The products formed are sodium sulfate and copper (II) hydroxide.

Copper (II) hydroxide forms a blue colored solution.

• sodium phosphate – Na₂HPO₂ copper sulfate – CuSO₄  

Na₂HPO₄ + CuSO₄ --> Na₂SO₄ + CuHPO₄

The products formed are sodium sulfate and copper (II) hydrogen phosphate.

Copper (II) hydrogen phosphate forms a blue colored solution.

• sodium chloride – NaCl silver nitrate – AgNO₃  

NaCl + AgNO₃--> AgCl + NaNO₃

The products formed are silver chloride and sodium nitrate.

Silver chloride forms a white precipitate.

• sodium hydroxide – NaOH silver nitrate – AgNO₃  

NaOH + AgNO₃   --> NaNO₃ + AgOH

The products formed are silver hydroxide and sodium nitrate.

Silver hydroxide forms a white precipitate.

• sodium phosphate – Na₂HPO₄ silver nitrate – AgNO₃

Na₂HPO₄ + AgNO₃  --> NaNO₃ +  Ag₂HPO₄

The products formed are sodium nitrate and silver hydrogen phosphate.

Silver hydrogen phosphate forms a colorless solution.

Explanation:

5 0
3 years ago
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