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3 years ago

If you were presented with 2l of a 2m sucrose stock solution, how many grams of sugar would be in a 100 ml aliquot?

1 answer:

4 0

The 2 L of sucrose stock solution would contain similar concentration with the 100 mL aliquot. Therefore the concentration of aliquot is still 2 M.

The molar mass of sucrose is 342.3 g / mol. Therefore the mass in a 100 mL (0.1 L) aliquot is:

mass = (2 mol / L) * 0.1 L * (342.3 g / mol)

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Reaction rate is expressed in terms of changes in the concentration of reactants and products. Write a balanced equation for the

KengaRu [80]

Answer : The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

$Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}$

The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

5 0

3 years ago

A 240g hydrated sodium sulphide contains 162g of water of crystallisation. What is the correct molecular formula for this compou

den301095 [7]

<span>A 240g hydrated sodium sulphide contains 162g of water of crystallisation. What is the correct molecular formula for this compound?
Answer is A

</span>

8 0

3 years ago

There are ________ σ bonds and ________ π bonds in h3c-ch2-ch=ch-ch2-c≡ch.

FrozenT [24]

The number of sigma and pi bonds are,

Sigma Bonds = 16

Pi Bonds = 3

Explanation:
Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.

6 0

3 years ago

This green solution of chromium(III) can further be reduced by zinc metal to a blue solution of chromium(II) ions. Write the bal

Contact [7]

Answer:

Half-reactions:

Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻

Net ionic equation:

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺

Explanation:

The Cr³⁺ is reduced to Cr²⁺:

<h3>Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>

Zn is oxidized to Zn²⁺:

<h3>Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>

Twice the reduction of Cr:

2Cr³⁺ + 2e⁻ → 2Cr²⁺

Now this reaction + Oxidation of Zn:

2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻

<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>

6 0

3 years ago

230 cm3 of hot tea at 96 °C are poured into a very thin paper cup with 100 g of crushed ice at 0 °C. Calculate the final tempera

ivann1987 [24]

Answer:

Final temperature of the ice tea, x = 42.73°C

Explanation:

230 cm^3 = 230 g of liquid ( taking the density of tea as the same the density of water)

In cooling from 96°C to 0°

C (before freezing) the available energy is

96 × 230 = 22080 cal

The latent heat (the heat required to melt the crushed ice per gram) of fusion for water = 79.7 cal/g

Melting 100 g of crushed ice at 0°C will absorb

100 × 79.7 = 7970 cal

The retain heat of the iced Tea = (22080 - 7970) cal of heat raising the mixture temperature as follows

(22080 - 7970) = 14110 cal

The mass of the final solution 230 + 100 = 330 g solution

Final temperature = 14110 / 330 = 42.76°C

Or where specific heat of water = 4.186J/Kg°C

and the latent heat of the ice = 334J/Kg

We have

230 × (specific heat capacity of water) × (96 - x) =100× (specific heat capacity of water) ×(0-x) + (latent heat) ×100 =

230 × 4.186 × (96 - x) =100× 4.186 ×(x-0) + 100×334

1381.38×x = 59026.88

Final temperature of the ice tea, x = 42.73°C

5 0

3 years ago