Organic protein catalysts are called:

How does Nathan’s hypothesis lead to new investigations

Simora [160]

Answer:

A) The data to support his hypothesis, so he should investigate if the same change happens in the density of solid water.

8 0

4 years ago

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The percent yield of this reaction is consistently 92.0%. CH4(g) + 4 S(g) → CS2(g) + 2 H2S(g) How many grams of sulfur would be

dybincka [34]

Answer:

132.17 g

Explanation:

The reaction given , in the question is -

CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )

From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .

since ,

Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂

Since ,

the Molecular weight of CS₂ = 76

Given , mass of CS₂ = 72.57 g

Moles of CS₂ = 72.57 / 76 = 0.95 mol

Since ,

The yield is 92.0 % .

Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles

Mass of S required = 4.13 * 32 = 132.17 g .

6 0

3 years ago

Hydrogen fluoride is used:

Simora [160]

Answer: The correct answer is option B)

in the synthesis of cryolite for aluminum production.

Explanation:

Hydrogen fluoride (HF) is used in synthesis of cryolite for aluminum production.

HF is a gas that dissolves in water to form hydrofluoric acid. However, HF also react with halogenated alkanes like trichloromethane CHCl3 to form organofluorine compounds that serves as precursors in the synthesis of cryolite, an Ore from which aluminum is produced

6 0

3 years ago

A gas occupies 525 mL at a pressure of 45.0 kPa. What would the volume of the gas be at a pressure of 65.0 kPa

choli [55]

Answer:

The volume of the gas at a pressure of 65.0 kPa would be 363 mL

Explanation:

Boyle's Law is a gas law that relates the pressure and volume of a certain amount of gas, without temperature variation, that is, at constant temperature.

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. In other words, the product P · V remains constant at the same temperature:

P*V=k

Being P1 and V1 the pressure and volume in state 1 and P2 and V2 the pressure and volume in state 2 are fulfilled:

P1*V1=P2*V2

In this case:

P1= 45 kPa= 45,000 Pa (being 1 kPa=1,000 Pa)
V1= 525 mL= 0.525 L (being 1 L=1,000 mL)
P2= 65 kPa= 65,000 Pa
V2= ?

Replacing:

45,000 Pa* 0.525 L= 65,000 Pa*V2

Solving:

$V2=\frac{45,000 Pa* 0.525 L}{65,000 Pa}$

V2=0.363 L=363 mL

The volume of the gas at a pressure of 65.0 kPa would be 363 mL

6 0

3 years ago

2. Answer the following questions about a sample of calcium phosphate:

weqwewe [10]

Answer:

a) 310.18 g/mol

b) 4.352 moles Ca3(PO4)2

c) 2.6 * 10^24 molecules

d) 5.24 * 10^24 P atoms

e)13.056 moles Ca

f)10825.3 grams

Explanation:

Step 1: Data given

Atomic mass of Ca = 40.08 g/mol

Atomic mass of P = 30.97 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molecular weight of Ca3(PO4)2

Molecular weight of Ca3(PO4)2 = 3*atomic mass of Ca + 2* atomic mass of P and 8* atomic mass of O

Molecular weight of Ca3(PO4)2 = 3*40.08 + 2*30.97 + 8*16.0 = 310.18 g/mol

Step 3: Calculate moles of Ca3(PO4)2 in 1350 grams

Moles Ca3(PO4)2 = mass Ca3(PO4)2 /molar mass

Moles Ca3(PO4)2 = 1350 grams / 310.18 g/mol

Moles Ca3(PO4)2 = 4.352 moles

Step 4: Calculate molecules in 1350 grams

Molecules = moles * number of Avogadro

Molecules = 4.352 moles * 6.02 * 10^23

Molecules = 2.6 *10^24 molecules

Step 5: Calculate moles Phosphorus

For 1 mol Ca3(PO4)2 we need 2 moles P

For 4.352 moles Ca3(PO4)2 we have 2*4.352 = 8.704 moles

Step 6: Calculate P atoms

Atoms P = 8.704 moles * 6.02*10^23

Atoms P = 5.24 * 10^24 P atoms

Step 7: Calculate moles Calcium in 1350 grams

For 1 mol Ca3(PO4)2 we have 3 moles Ca

For 4.352 moles we have 3*4.352 = 13.056 moles Ca

Step 8: Calculate mass of 2.1 * 10^25 molecules of Ca3(PO4)2

Moles Ca3(PO4)2 = 2.1 * 10^25 / 6.02 * 10^23

Moles Ca3(PO4)2 = 34.9 moles

Mass Ca3(PO4)2 = 34.9 moles * 310.18 g/mol

Mass Ca3(PO4)2 = 10825.3 grams

4 0

3 years ago