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4 years ago

Help quickly will give brainliestt

Chemistry

1 answer:

galben [10]4 years ago

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Answer:

Diffusion is driven by differences in concentration. When chemical substances such as perfume are let loose in a room, their particles mix with the particles of air. Diffusion in gases is quick because the particles in a gas move quickly. It happens even faster in hot gases because the particles of gas move faster.

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

Which pure substance can be classified as an element?

taurus [48]

Watər H20 because it is formed by the process of hydrogen and oxygen bonding

5 0

3 years ago

Read 2 more answers

Which of the following elements is the least electronegative?

Pepsi [2]

Answer:

Explanation:

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8 0

3 years ago

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What is the value for ΔS°reaction for the following reaction, given the standard entropy values

QveST [7]

Standard entropy of the compounds of interest are as follows,
$S^{0} (C6H12O6)$ = 212.1 J/K.mol
$S^{0} (O2)$ = 205.0 J/K.mol
$S^{0} (CO2)$ = 213.6 J/K.mol
$S^{0} (H2O)$ = 69.9 J/K.mol

Now for the reaction:
C6H12O6(s) + 6O2(g) —->6CO2(g) + 6H2O(l)

ΔS°reaction = ∑ $S^{0} products$ - ∑ $S^{0} reactants$
∴ ΔS°reaction =( 6 $S^{0} (CO2)$ + 6 $S^{0} (H2O)$ ) - ( $S^{0} (C6H12O6)$ + 6 $S^{0} (O2)$ = 205.0 J/K.mol)
∴ ΔS°reaction = [6 (213.6) + 6 (69.9 )] - [(212.1) + 6(205.0)]
∴ ΔS°reaction = 258.9

5 0

3 years ago

Read 2 more answers

Which choice best describes the specific heat of water compared with that of metals?

n200080 [17]

Answer: C. The specific heat of water is greater than the specific heat of metals.

The specific heat is defined as the amount required to raise the temperature of a unit mass of a substance by 1 degree Celsius.

This is expressed mathematically as

Q= mc∆T

Where Q is the energy/heat required which is measured in Joules.

m is the mass (grams)

c is the specific heat which is measured in joule/gram degree Celsius.

∆T- change in temperature

Substance which has a high specific heat require a lot of heat for its temperature to be raised by one degree. On the other hand substances with lower specific heat require only little amount of heat for its temperature to be raised by one degree.

Consider an equal mass of metal and water. If both are heated at the same time,the metal would become hotter than the water much faster. This is because the specific heat of the metal is lower than the water. Hence it requires only a little heat for its temperature to raised by one degree.

Thus we can conclude that the specific heat of water is much greater than that of a metal.

3 0

3 years ago