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2 years ago

6

in which sample of water do the molecules have the highest average kinetic energy 1)20. mL at 100.°C 2)40. mL at 80. °C 3)60. mL

at 60.° C 4)80. ml at 40.° C

1 answer:

miv72 [106K]2 years ago

6 0

100°C because all the molecules are moving the fastest past each other

You might be interested in

Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

<em>Complete text:</em>

<em>Honors Stoichiometry Activity WorksheetInstructions: </em>

<em>Activity Two: Just Lemons, Inc. Production</em>

<em>Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.</em>

<em>Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. </em>

<em>Water 946.36 g </em>

<em>Sugar 196.86 g </em>

<em>Lemon Juice 193.37 g </em>

<em>Lemonade 2050.25g</em>

<em>Leftover Ingredients?</em>

<em>Just Lemons Lemonade Recipe Equation:</em>

<em>2 water + sugar + lemon juice = 4 lemonade</em>

<em>Mole conversion factors:</em>

<em>1 mole of water = 1 cup = 236.59 g</em>

<em>1 mole of sugar = 1 cup = 225 g</em>

<em>1 mole of lemon juice = 1 cup = 257.83 g</em>

<em>1 mole of lemonade = 1 cup = 719.42 g</em>

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

<u>Water:</u>

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

<u>Sugar:</u>

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

<u>Lemon juice:</u>

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

7 0

2 years ago

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

vesna_86 [32]

(a) In this section, give your answers to three decimal places.

(i)

Calculate the mass of carbon present in 0.352 g of CO

2

.

Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g

of

A

.

(ii)

Calculate the mass of hydrogen present in 0.144 g of H

2

O.

Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g

of

A

.

(iii)

Use your answers to calculate the mass of oxygen present in 0.240 g of

A

Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g

of

A

(b)

Use your answers to

(a)

to calculate the empirical formula of

A

thank you

hope it helpsss

4 0

2 years ago

How many milliliters of 0.02 M HCl are needed to react completely with 100 mL of 0.01 M NaOH?

serious [3.7K]

The reaction of HCl and NaOH is HCl + NaOH = NaCl + H2O. So the mole number of HCl and NaOH is equal. So the volume of HCl =0.01*0.1/0.02=0.05 L =50 ml. So the answer is D).

3 0

2 years ago

Read 2 more answers

Which of the following species contains manganese with the highest oxidation number?

ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0

1 year ago

If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the

jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S

1 g of NH₃ will produce $\frac{68}{34}$ g of (NH₄)₂S

25g of NH₃ will produce $\frac{65}{34} X 25 g$ of (NH₄)₂S

= 47.8 g of (NH₄)₂S

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S

4 0

2 years ago