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Charra [1.4K]
3 years ago
13

Many popular beverages are sold in two-kilo bottles. true or false

Chemistry
1 answer:
Molodets [167]3 years ago
5 0
False. The answer is liter. kilo is the base (liter) times 1000. it would make no sense.
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A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.130 M AuNO 3 solution at 25 °C. The gold wire i
padilas [110]

Answer:

The half reaction occurring at this electrode will be given as:

Au^++e^-\rightarrow Au

Explanation:

In electrochemical cell reduction occurs at anode.

But the gold wire is connected to the positive terminal of the battery which means that gold wire will act as a cathode nad reduction will take place

The half reaction occurring at this electrode will be given as:

Au^++e^-\rightarrow Au

7 0
3 years ago
What is the overall trend for ionization energy on the periodic chart?
gizmo_the_mogwai [7]

Answer:

The general trend is for ionization energy to increase moving from left to right across an element period. Moving left to right across a period, atomic radius decreases, so electrons are more attracted to the (closer) nucleus.

Explanation:

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3 years ago
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Is crumpling a piece of paper chemical changes
kenny6666 [7]
No, because you are not changing the chemical make-up of the paper
8 0
3 years ago
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Molybdenum can form a wide series of halide compounds, including four different fluoride compounds. The percent by mass of molyb
azamat

The formula and names of the compounds are:

1. Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride  

4. Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride  

1. Determination of the name and formula of the molybdenum fluoride having 63.0% of molybdenum.

Molybdenum (Mo) = 63.0%

Fluorine (F) = 100 – 63 = 37%

<h3>Formula =? </h3>

Mo = 63.0%

F = 37%

Divide by their molar mass

Mo = 63.0 / 96 = 0.656

F = 37 / 19 = 1.947

Divide by the smallest

Mo = 0.656 / 0.656 = 1

F = 1.947 / 0.656 = 3

Therefore,

Formula of compound => MoF₃

Name of compound => Molybdenum trifluoride

2. Determination of the name and formula of the molybdenum fluoride having 56.0% of molybdenum.

Molybdenum (Mo) = 56.0%,

Fluorine (F) = 100 – 56 = 44%

<h3>Formula =? </h3>

Mo = 56%

F = 44%

Divide by their molar mass

Mo = 56 / 96 = 0.583

F = 44 / 19 = 2.316

Divide by the smallest

Mo = 0.583 / 0.583 = 1

F = 2.316 / 0.583 = 4

Therefore,

Formula of compound => MoF₄

Name of compound => Molybdenum tetrafluoride

3. Determination of the name and formula of the molybdenum fluoride having 50.0% of molybdenum.

Molybdenum (Mo) = 50.0%,

Fluorine (F) = 100 – 50 = 50%

<h3>Formula =? </h3>

Mo = 50%

F = 50%

Divide by their molar mass

Mo = 50 / 96 = 0.520

F = 50 / 19 = 2.632

Divide by the smallest

Mo = 0.520 / 0.520 = 1

F = 2.632 / 0.520 = 5

Therefore,

Formula of compound => MoF₅

Name of compound => Molybdenum pentafluoride  

4. Determination of the name and formula of the molybdenum fluoride having 46.0% of molybdenum.

Molybdenum (Mo) = 46.0%,

Fluorine (F) = 100 – 46 = 54%

<h3>Formula =? </h3>

Mo = 46%

F = 54%

Divide by their molar mass

Mo = 46 / 96 = 0.479

F = 54 / 19 = 2.842

Divide by the smallest

Mo = 0.479 / 0.479 = 1

F = 2.842 / 0.479 = 6

Therefore,

Formula of compound => MoF₆

Name of compound => Molybdenum hexafluoride  

Learn more: brainly.com/question/11185156

7 0
2 years ago
What is the quantity of atoms in KMnO4 ?
tankabanditka [31]

Answer:

Potassium permanganate has a molar mass of 158.04 g/mol. This figure is obtained by adding the individual molar masses of <em><u>four oxygen atoms</u></em>, <em><u>one manganese atom</u></em> and <em><u>one potassium atom</u></em>

Explanation:

6 0
3 years ago
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