All categories

3 years ago

This is a steel penny from 1943. A steel penny will weigh 2.72 g and is

Chemistry

1 answer:

ehidna [41]3 years ago

5 0

Answer:

The easiest way to determine if a 1943 cent is made of steel, and not copper, is to use a magnet. If it sticks to the magnet, it is not copper. If it does not stick, the coin might be of copper and should be authenticated by an expert

Explanation:

You might be interested in

Use the following equation to answer the questions below.

notsponge [240]

Answer:

202

Explanation:

7 0

2 years ago

each element is designated by its __________ which is usually from the first letters of the elements name

klemol [59]

Answer:

each element is designated by its <u>Chemical Symbol </u>which is usually from the first letters of the elements name

Explanation:

7 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

2 years ago

Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.75 L of a 0.130 M sucro

Advocard [28]

Answer:125.84g

Explanation:Sucrose is dissacharides an organic compound in the class of carbonhydrate with the chemical formula C11H22O11.molar concentration is given by number of moles/Volume,this implies that moles=molar concentration ×Volume=0.130M×2.75L=0.3575moles.

Furthermore,number of moles=Mass of Sucrose/molecular Mass of Sucrose.

From it's formular C11H22O11, molecular Mass is the addition of the mass number which is 12 for C,2 for H and 16 for oxygen,O.so molecular Mass of Sucrose is (12×11)+(2×22)+(16×11)=352.

So mass =moles ×molecular mass=0.3575moles×352g/moles=125.84g

3 0

2 years ago

An atom of boron has an atomic number of 5 and an atomic mass of 11. the atom contains

kirill [66]

The atom contain 6 neutrons, if that's the question

8 0

3 years ago