Ask question

Ask question

All categories

3 years ago

6

Use numbers to indicate the order of the steps in the titration process.

2 answers:

Snowcat [4.5K]3 years ago

7 0

Answer:

7

5

1

8

3

6

2

4

just took on ed2020, good luck

OleMash [197]3 years ago

4 0

Answer:

7

5

1

8

3

6

2

4

Explanation:

You might be interested in

3- AR 385-63/MCO 3570.1C is used in conjunction with

Rasek [7]

AR 385-63/MCO 3570.1C is used in conjunction with DA PAM 385-63.
AR 385-63/MCO 3570.1C is the regulation or order which provides revised range safety policy for the Army and Marine Corps.
This order/regulation applies on the Active Army or the Army National Guard of the United states.

8 0

3 years ago

What is the temperature in kelvins of -14°C?

julia-pushkina [17]

Answer choice B 259 kelvins

3 0

2 years ago

Match the terms with their correct definition:

geniusboy [140]

<h2><em>1. A</em></h2><h2><em>3. B</em></h2><h2><em>4. C</em></h2><h2><em>7. E</em></h2><h2><em>5. F</em></h2>

7 0

3 years ago

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

Along with subduction, what other process changes sedimentary rocks to metamorphic rocks?

muminat

It would be weathering because of all the heat and pressure.

7 0

3 years ago