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4 years ago

Temperature Scale Conversion

Chemistry

1 answer:

earnstyle [38]4 years ago

3 0

Answer: b. -40 ° F , 233 K

c. 1947.2 °F , 1337 K

Explanation: To convert degree Celsius to Fahrenheit we will use the following formula.

°F = 9/5 x ( °C ) + 32

= 9/5 x (-40.0 °C ) + 32 = -40 °F

°F = 9/5 x ( °C ) + 32

= 9/5 x (1064 °C ) + 32 = 1947.2 °F

To solve for Kelvin just add 273 in °C.

-40.0°C + 273 = 233 K

1064 °C + 273 = 1337 K

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Answer:

See explanation

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-) True or False: All R stereocenters are dextrorotatory.

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-) True or False: Normal linear amines can be chiral centers.

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-)True or False: If the R isomer of a molecule has an optical rotation of + 25.7 degree then the S isomer of the molecule will have an optical rotation of -25.7 degree.

If we have the exact opposite (as we have in this case) the magnitud of the optical activity value will remain and the sign will change. TRUE

-)True or False: A CN is a higher priority than a CH2OH.

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-)True or False: A CH2CH2Br is higher priority than a CH2F.

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4 years ago

You need to make an aqueous solution of 0.207 M calcium acetate for an experiment in lab, using a 300 mL volumetric flask. How m

jeka94

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

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3 years ago

A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta

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Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole

The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited

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3 years ago