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3 years ago

How are tectonic plates related to earthquakes and volcanoes?

Chemistry

1 answer:

patriot [66]3 years ago

8 0

Answer

Tectonic plates are parts of the earth's crust. There are cracks in these plates that allow magma to seep through and build-up which creates volcanic eruptions. Volcanos themselves are created by the earth's plates pushing against each other and slowly, over time, rising to create these large raises in the crust. Earthquakes are caused by these plates creating friction against one another until they forcefully overlap or rip apart, hence the shaking effect during an earthquake.

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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 atom below would be MOST likely to form a compound with magnesium (Mg) in a ratio of one to one?

fiasKO [112]

Answer:

Explanation:

5 0

3 years ago

PLEEAASSEE HEELLPP MEE I NEED THIS ANSWER TO FINISH THE LAST QUESTION!!!!

VARVARA [1.3K]

The product label should be at the bottom hopes this helped

3 0

3 years ago

Read 2 more answers

A solution has a hydrogen ion concentration of 6.8 10–3 M. (2 points) i. What is the pH of the solution? Show your work.

vladimir2022 [97]

Answer:

2.17

Explanation:

To calculate pH using the hygrogen ion concentration we must use the following formula:

- log (H+) = pH

All we have to do now is plug in our hygrogen ion concentration and put it in our calculator.

- log (6.8 × 10⁻³) = 2.17

Don't forget proper figures for pH!

8 0

3 years ago

A balloon containing helium gas expands from 230

Anit [1.1K]

The answer for the following problem is mentioned below.

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

Explanation:

Given:

Initial volume ( $V_{1}$ ) = 230 ml

Final volume ( $V_{2}$ ) = 860 ml

Initial moles ( $n_{1}$ ) = 3.8 × $10^{-4}$ moles

To find:

Final moles ( $n_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

V ∝ n

$\frac{V_{1} }{V_{2} }$ = $\frac{n_{1} }{n_{2} }$

where,

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $n_{1}$ ) represents the initial moles of the gas

( $n_{2}$ ) represents the final moles of the gas

Substituting the above values;

$\frac{230}{860}$ = $\frac{3.8 * 10^-4}{n_{2} }$

$n_{2}$ = 14.2 × $10^{-4}$ moles

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

7 0

3 years ago