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2 years ago

How many grams of water would there be in 100.0g of hydrate? How many moles?

2 answers:

7 0

Answer: The mass of water present in given amount of hydrate is 36.68 grams and number of moles of water are 2.04 moles.

Explanation:

We are given:

Mass of hydrate = 0.946 grams

Mass of water present = 0.347 grams

We need to calculate the mass of water present in 100 grams of hydrate. By using unitary method, we get:

In 0.946 g of hydrate, the amount of water present is 0.347 g

So, in 100 g of hydrate, the amount of water present will be = $\frac{0.347g}{0.946g}\times 100g=36.68g$

Hence, the mass of water present in given amount of hydrate is 36.68 grams.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 36.68 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{36.38g}{18g/mol}=2.04mol$

Hence, the number of moles of water are 2.04 moles.

IceJOKER [234]2 years ago

6 0

The calculation for the amount of water present in the given amount of hydrate is shown below,
amount water = (100 g hydrate) x (0.347 g H2O / 0.946 g hydrate)
= 36.68 g
Thus, the amount of water present in the hydrate is approximately 36.68 g.

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Which statement best applies collision theory to preventing a dangerous reaction from occurring? Store the reactants together at

olasank [31]

Answer:

According to collision theory, to preventing a dangerous reaction from occurring, the best is to keep reactants in separate containers (last statement).

Justification:

Collison theory states that the the reactant substances (atoms, ions or molecules) must first collide to react and form the products.

Additionally, to form the products, the collisions must meet two requirements:

the reactant substances must collide with the correct orientation, and

the reactant substances must collide with energy enough to form the activated complex (transition state).

Hence, the collision theory permits you preventing a dangerous reaction from occurring, by using the elemental knowledge that the substances must first collide in order to they react, and so the most effective way is to keep the reactants in separate contaners, preventing the reactants from coming into direct contact.

7 0

2 years ago

Read 2 more answers

Substitution of an amino group on the para position of acetophenone shifts the cjo frequency from about 1685 to 1652 cm−1 , wher

cluponka [151]

Answer:

Here's what I get.

Explanation:

The frequency of a vibration depends on the strength of the bond (the force constant).

The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.

Acetophenone

Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.

p-Aminoacetophenone

The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.

The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.

p-Nitroacetophenone

The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.

As electron density moves toward C-1, the double bond character of the C=O group increases.

The bond length decreases, so the bond becomes stronger, and wavenumber increases to 1693 cm¹.

6 0

2 years ago

Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

stira [4]

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the lower pKa value and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine

8 0

2 years ago

After passing through pyruvate dehydrogenase and the citric acid cycle, one mole of pyruvate will result in the formation of ___

mamaluj [8]

Answer:

The answer to be filled in the respective blanks in question is

3 and 1

Explanation:

So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.

Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1

7 0

2 years ago

By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm

kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

so for 1 quart = 0.943 L

$\tt 0.943~L=0.943\times 10^{-3}m^3$

Volume of oil dumped = volume of swimming pool

$\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}$

3 0

2 years ago