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lora16 [44]
3 years ago
10

The boiling points, at standard pressure, of four compounds are given in the table below. Which type of attraction can be used t

o explain the unusually high boiling point of H2O?
Chemistry
1 answer:
Phantasy [73]3 years ago
7 0

Explanation:

Hydrogen bonding! Each hydrogen atom with a high charge density forms bonds with lone pairs on oxygen atoms. These bonds require a lot of energy to break giving water a relatively high boiling point.

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Explain why the ability of PLP to catalyze an amino acid transformation is greatly reduced if the OHOH substituent of pyridoxal
cupoosta [38]

Answer:

Following are the responses to the given question:

Explanation:

In the given question the OH could be generated as the H-bond with both the N (nitrogen), which is used to place its N as the partially positive value(+). It is used to simplifies the addition for either its  AA with the imine C. In the above H-bond, the w N is not required as the possibility to use the OCH3

7 0
3 years ago
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
4 years ago
a 5-meter ramp lifts objects to a height of 0.75 meters.What is the mechanical advantage of the ramp?
coldgirl [10]
This is an Engineering question, and the answer is reasonably simple. The mechanical advantage is given by:

MA = \frac{Distance 1}{Distance 2} on a lever or ramp system.

MA = \frac{5}{0.75}

MA = 6.67 (there are no units, as MA is a ratio)


5 0
4 years ago
For the weak acid ch3cooh (acetic acid) that is titrated with a strong base (naoh), what species (ions/molecules) are present in
liq [111]
This is a strong base / week acid reaction.

NaOH + CH3COOH

The equilibrium of this reaction is very displaced to the right leading to the formation of the products

Na CH3COO + H2O

Na CH3COOH is a ionic compound which in solutionn will be as Na (+) and CH3COOH(-)

=> CH3COOH + NaOH = CH3 COO(-) + Na(+) + H2O

So, the predominant species in the solution are the ions Na(+) and CH3COO(-).

In general, in an strong base / weak acid titration, the predominant species present at the stoichiometric point will be the cation of the strong base (Na+ in this case) and the conjugate base of the weak acid (the anion of the weak acid, which is CH3COO- in this case).

The answer is predominantly Na(+) and CH3COO(-); predominantly because it is an equlibrium which means that the rectants will also br present.
3 0
3 years ago
Please answer ASAP!!
Alenkasestr [34]

Answer:

C.0.28 V  

Explanation:

Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:

The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:

Cu²⁺ + 2e⁻ → Cu E° = 0.52V

Ag⁺ + 1e⁻ → Ag E° = 0.80V

As the Cu will be oxidized:

Cu → Cu²⁺ + 2e⁻

The cell potential is:

E°Cell = E°cathode(reduced) - E°cathode(oxidized)

E°cell = 0.80V - (0.52V)

E°cell = 1.32V

Right answer is:

<h3>C.0.28 V </h3>

<h3 />

4 0
3 years ago
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