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2 years ago

Which unit represents 100 meters?

Chemistry

2 answers:

adell [148]2 years ago

6 0

Answer:

Hectometer.

Explanation:

<u><em>Hecto- means 100; a hectometer is 100 meters.</em></u>

Oduvanchick [21]2 years ago

4 0

<h2>Answer:</h2>

<u>The unit represents 100 meters is called </u><u>hectometer</u>

<h2>Explanation:</h2>

The hectometer is not commonly used unit of length in the metric system. This unit is equal to one hundred metres.

The word hectometers comes from a combination of two words "metre" which is unit of length and the SI prefix "hecto-" which means "hundred".

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Sodium (chemical symbol Na) is atomic number 11 and is in the first column of the periodic table so it has one valence electron.

lisabon 2012 [21]

Na is cation so it loses electron to be positive and become stable losing one valence shells one electron so it's oxidation number is +1 ie A is correct

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3 years ago

Read 2 more answers

The following reaction has an equilibrium constant (Keq) of 1 under standard conditions and can be catalysed by the enzyme aspar

Trava [24]

Answer:

What is the AGⓇ of this reaction? 0.
Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.

Explanation:

1. To calculate the delta G of a reaction given the K, we use the following equation:

ΔG°= -RT ln K.

Which gives us 0 when K is 1.

2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.

3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.

4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.

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2 years ago

HELP please! Which of these describes how heat is transferred by conduction

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D. Air molecules touch the warm ground, heating them up

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2 years ago

If you mix water with your urine drug test will you pass

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2 years ago

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

2 years ago