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3 years ago

A) 24 g B) 36 g C) 48 g D) 60 g

Chemistry

1 answer:

Hitman42 [59]3 years ago

5 0

Answer:

I'm not sure if this right because i've only taken one year of Chemistry but the answer I got was

A. 24 g

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43 milliliters of water weighs 43 g. what is the density of the water?

Anastaziya [24]

Answer:

$\rho =1g/mL$

Explanation:

Hello,

In this case, since the density is defined as the ratio between the mass and the volume as shown below:

$\rho =\frac{m}{V}$

We can compute the density of water for the given 43 g that occupy the volume of 43 mL:

$\rho =\frac{43g}{43mL}=1g/mL$

Regards.

4 0

3 years ago

Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

Answer: The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

Explanation:

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

4 years ago

How many grams of O2(g) are needed to completely burn 15.6 g of C3H8(g)?

alexdok [17]

15.6gC3H8*44.1gC3H8*32gO2=11.32

7 0

4 years ago

How many grams are in 2.45E24 formula units of Fp3BZ2? The molar mass of Fp3Bz2. is 97.05 g/mol.

dlinn [17]

Answer:

394.99g

Explanation:

The number of moles of a substance can be calculated by dividing the number of atoms of such substance by Avagadro's number (6.02 × 10^23)

n = nA ÷ 6.02 × 10^23

The number of atoms of Fp3BZ2 in this question is 2.45E24 formula units i.e. 2.45 × 10^24

n = 2.45 × 10^24 ÷ 6.02 × 10^23

n = 2.45/6.02 × 10^(24-23)

n = 0.407 × 10¹

n = 4.07moles

Using mole = mass/molar mass

Where; molar mass of Fp3Bz2. is 97.05

g/mol.

mass = molar mass × mole

mass = 97.05 × 4.07

mass = 394.99g

4 0

3 years ago

Which label belongs in the region marked X?

lubasha [3.4K]

Answer:

A) involves changes in temperature

Explanation:

The figure is missing, but I assume that the region marked X represents the region in common between Gay-Lussac's law and Charle's Law.

Gay-Lussac's law states that:

"For an ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as

$p\propto T$

where p is the pressure of the gas and T its absolute temperature.

Charle's Law states that:

"For an ideal gas kept at constant pressure, the volume of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as

$V\propto T$

where V is the volume of the gas and T its absolute temperature.

By looking at the two descriptions of the law, we see immediately that the property that they have in common is

A) involves changes in temperature

Since the temperature is NOT kept constant in the two laws.

7 0

3 years ago