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3 years ago

Does anyone know the anserw the symbol of the element

Chemistry

1 answer:

Komok [63]3 years ago

3 0

Cesium.

Groups are the vertical columns that run up and down while periods are the horizontal rows. So to find the answer to this, go to the first column (Group 1) and find the sixth period (row 6) which will land you on Cesium, element 55.

Hope this helps!

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Which rule for assigning oxidation numbers is correct?

LenKa [72]

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Oxygen is usually –2. This is CORRECT.
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A monatomic ion is 0. This is INCORRECT. Diatomic ion is 0.

4 0

3 years ago

Read 2 more answers

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o

jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

Step 1: The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

Step 2: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

Step 3: Calculate moles of glucose

Moles glucose = Mass glucose / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

Step 4: Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

Step 5: Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

Step 6: Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

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3 years ago

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Answer:

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Mark me as brainilist pls

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2 years ago