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3 years ago

2. What is an example of a molecule with different types of atoms?

Chemistry

1 answer:

agasfer [191]3 years ago

4 0

Answer:

a compound

Explanation:

A molecule with multiple atoms is compound like water it's made up of 1 oxygen and two hydrogen.

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Determine the total number of atoms contain in a 2.00 moles of Ni

spayn [35]

Answer:

2.00 moles of Ni has 1.2 *10^24 atoms

Explanation:

Step 1: Data given

Number of moles Ni = 2.00 moles

Number of Avogadro = 6.022*10^23 /mol

Step 2: Calculate number of atoms

Number of particles (=atoms) = Number of Avogadro * number of moles

Number of atoms = 6.022 * 10^23 /mol * 2.00 moles

Number of atoms = 1.2*10^24 atoms

2.00 moles of Ni has 1.2 *10^24 atoms

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3 years ago

Determination of your estimate determines the BLANK and BLANK of a piece of equipment.

Zina [86]

Answer:

scalpel,mask

Explanation:

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2 years ago

Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W

alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

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3 years ago

A common misconception people have is that sugar is energy and energy is sugar but you know better. Create a clear explanation t

Gnom [1K]

Answer:There are important differences between naturally occurring sugars, ... Understanding these differences can help you make choices that are better for your health. ... of energy, along with a bevy of vitamins, minerals, and phytonutrients. ... sugar substitutes, but many people find they have an unpleasant

Explanation:

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2 years ago

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

4 0

3 years ago