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3 years ago

Heat from a fire is an example of? a Radiation b Conduction c Convection

Chemistry

2 answers:

nikdorinn [45]3 years ago

7 0

The answer should be A because radiation refers to the emission of energy in rays or waves

Nat2105 [25]3 years ago

4 0

I think the answer is A bit I could be wrong

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The element is abundant in earths crust and combines with oxygen to form rocks and minerals. what element is it?

romanna [79]

The element is silicon (Si). It forms minerals like quartz and rocks such as granite.

5 0

3 years ago

13.8g element consist 4.6 ×10^22 atoms what is atomic mass of element?

Keith_Richards [23]

<h2>~Solution :-</h2>

Here, to find the atomic mass of element, we must;

We know that,

4.6 x $ \sf{10^{22}}$ atoms of an element weigh 13.8g.

Thus,

The atoms of $ \sf{ 6.02 \times 10^{13}}$ will weigh;

$= \sf{\frac{13.8 \times 6.02 \times {10}^{23} }{4.6 \times {10}^{22} } } \\$

$= \sf{180.6 \: g \: \: (molar \: mass)}$

Hence, the molar mass (atomic mass) will be 180.6 g.

8 0

2 years ago

Read 2 more answers

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

3 years ago

12a. How does the model of the volcano provide evidence for the cycling of Earth materials through the rock cycle?

Bond [772]

Answer:

lava is coming from the crust

Explanation:

volcano's explode with lava because the crust pushes it out

6 0

2 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of

Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

No of moles = mass of the substance÷molar mass of the substance

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

Thus 52.8 g of CO2 is produced.

5 0

3 years ago