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3 years ago

What is the molarity of a solution where there are 1.35 moles of H2SO4 in 3.00 L of solution?

Chemistry

1 answer:

Montano1993 [528]3 years ago

4 0

Answer: 0.450 M H2SO4

Explanation:

To solve this problem, we must remember how to compute molarity. To find the molarity of a solution, we divide the number of moles by the number of liters of solution. Using this formula and substituting the given values, we get:

Molarity = moles solute/liters solution

= 1.35 moles/3.00 L

= 0.450 M H2SO4

Therefore, the correct answer is 0.450 M H2SO4. Note that the answer has 3 significant figures because each of the given values also contains 3 significant figures.

Hope this helps!

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a 55g block of metal has a specific heat of 0.45 J/g°C. What will be the temperature change of this metal if 450 J of heat energ

Dafna11 [192]

Answer:

18.18C

Explanation: Write whats below to show your work :D

m=55

C=0.45

Q=450zj

ch.temp=?

Q=mct

450 = (55)(.45)T

450 = 24.75

/24.75 = /24.75 (cancels out)

18.18 is the answer

Hope this helps! did it with my teacher.

7 0

3 years ago

What is tins atomic symbol

qaws [65]

Tin
Chemical Element
Tin is a chemical element with the symbol Sn and atomic number 50. It is a main group metal in group 14 of the periodic table. Wikipedia
Symbol: Sn
Electron configuration: [Kr] 4d105s25p2
Atomic number: 50
Melting point: 449.5°F (231.9°C)
Atomic mass: 118.71 u
Boiling point: 4,717°F (2,603°C)
Electrons per shell: 2, 8, 18, 18, 4

4 0

4 years ago

Read 2 more answers

How many kilograms of iron can be obtained from 100 kilograms of Fe203

ANTONII [103]

Answer:

754

Explanation:

5 0

3 years ago

Identify the type of bonding within each substance. Co ( s ) ionic covalent metallic CoCl 2 ( s ) covalent ionic metallic CCl 4

krok68 [10]

Answer:

1. Co ( s ) - metallic bonding

2. CoCl₂ ( s ) - ionic bonding

3. CCl₄ ( l ) - covalent bonding

Explanation:

Metallic bonding -

It is the type of bonding present between the atoms of the metals , via the electrostatic interaction between the metal and the delocalized electrons , is known as metallic bonding .

For example ,

Mostly metals show metallic bonding .

Ionic bonding -

It is the type of bonding present between the ions i.e. , the cation and the anion is known as ionic bonding .

For example ,

Mostly ionic compound , like salts show ionic bonding .

Covalent bonding -

It is the type of bonding which is present between shared pair of electrons , is known as covalent bonding .

For example ,

Most of the carbon compounds are capable to show covalent bonding .

Hence , from the question ,

1. Co ( s ) - metallic bonding

2. CoCl₂ ( s ) - ionic bonding

3. CCl₄ ( l ) - covalent bonding

6 0

3 years ago

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin pla

RSB [31]

<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 25.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g$

To calculate the heat released by the reaction, we use the equation:

$q=mc\Delta T$

where,

q = heat released

m = Total mass = [1.25 + 25] = 26.25 g

c = heat capacity of water = 4.18 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(21.9-25.8)^oC=-3.9^oC$

Putting values in above equation, we get:

$q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ$

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

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

Given mass of ammonium nitrate = 1.25 g

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

$\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -0.428 kJ

n = number of moles = 0.0156 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol$

Hence, the enthalpy change of the reaction is -27. kJ/mol

4 0

4 years ago