CHEMISTRY HIGH SCHOOL

"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the net flow of energy transferred to the environment in 1 second?"

Answers

Answer 1

Answer:

Explanation:

It is known that formula for area of a sphere is as follows.

A = $4 \pi r^(2)$

= $4 * 3.14 * (0.50 m)^(2)$

= 3.14 $m^(2)$

$T_(a)$ = (27 + 273.15) K = 300.15 K

T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

Q = $esA(T^(4) - T^(4)_(a))$

where, e = emissivity = 0.85

s = stefan-boltzmann constant = $5.6703 * 10^(-8) Wm^(-2) K^(-4)$

A = surface area

Hence, putting the given values into the above formula as follows.

Q = $esA(T^(4) - T^(4)_(a))$

= $0.85 * 5.6703 * 10^(-8) Wm^(-2) K^(-4) * 3.14 * ((350.15)^(4) - (300.15)^(4))$

= 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

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What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter? NaOH (s) → Na+ (aq) + OH– (aq) ∆H = -44.5 kJ/mol

Answers

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=(q)/(n)$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=(1.52g)/(40g/mole)=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=(q)/(0.038mol)$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m* c* (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g* 4.18J/g^oC* (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

A certain element has a melting point over 700 ∘C and a density less than 2.00 g/cm3. What is one possible identity for this element?

Answers

The only answer to an element that melts over 700 degrees Celcius and has a density of less than 2 g/cm3 is Beryllium. This element is rarely found anywhere in the entire universe. It's atomic number is 4 and it is created through stellar nucleosynthesis.

The element is Beryllium

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3 ( s ) + 2 HCl ( aq ) ⟶ CaCl2 ( aq ) + H 2 O ( l ) + CO 2 ( g ) How many grams of calcium chloride will be produced when 30.0 g of calcium carbonate is combined with 10.0 g of hydrochloric acid

Answers

Answer: The mass of $CaCl_2$ produced is, 15.2 grams.

Explanation : Given,

Mass of $CaCO_3$ = 30.0 g

Mass of $HCl$ = 10.0 g

Molar mass of $CaCO_3$ = 100 g/mol

Molar mass of $HCl$ = 36.5 g/mol

First we have to calculate the moles of $CaCO_3$ and $HCl$ .

$\text{Moles of }CaCO_3=\frac{\text{Given mass }CaCO_3}{\text{Molar mass }CaCO_3}$

$\text{Moles of }CaCO_3=\frac{\text{Given mass }CaCO_3}{\text{Molar mass }CaCO_3}=(30.0g)/(100g/mol)=0.300mol$

and,

$\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}$

$\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}=(10.0g)/(36.5g/mol)=0.274mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $HCl$ react with 1 mole of $CaCO_3$

So, 0.274 moles of $HCl$ react with $(0.274)/(2)=0.137$ moles of $CaCO_3$

From this we conclude that, $CaCO_3$ is an excess reagent because the given moles are greater than the required moles and $HCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CaCl_2$

From the reaction, we conclude that

As, 2 mole of $HCl$ react to give 1 mole of $CaCl_2$

So, 0.274 mole of $HCl$ react to give $(0.274)/(2)=0.137$ mole of $CaCl_2$

Now we have to calculate the mass of $CaCl_2$

$\text{ Mass of }CaCl_2=\text{ Moles of }CaCl_2* \text{ Molar mass of }CaCl_2$

Molar mass of $CaCl_2$ = 110.98 g/mole

$\text{ Mass of }CaCl_2=(0.137moles)* (110.98g/mole)=15.2g$

Therefore, the mass of $CaCl_2$ produced is, 15.2 grams.

How many hydrogen atoms are in the following molecule of ammonium sulfide? (NH4)2Sa. 2
b. 4
c. 8
d. 16

Answers

There are eight hydrogen atoms in the following molecule of ammonium sulfide. Thus, the correct option for this question is C.

What is a Molecule?

A molecule may be defined as a cluster of atoms that are significantly bonded together in order to represent the smallest basic unit of a chemical compound that can involve in a chemical reaction.

According to this question, the correct representation of ammonium sulfide is $(NH_4)_2S$ , where the number of the hydrogen atoms is equal to 4 × 2 = 8. Apart from this, the number of nitrogen atoms is 2. While the number of sulfur atoms is only 1.

Therefore, there are eight hydrogen atoms in the following molecule of ammonium sulfide. Thus, the correct option for this question is C.

To learn more about Molecules, refer to the link:

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The are eight hydrogen atoms in ammonium sulfide because there are 2 molecules of ammonium.

60 ml of a 0.40 m solution of h2so4 is used to neutralize 0.3 m magnesium hydroxide solution, what is the ml of sodium hydroxide?

Answers

60 ml of a 0.40 m solution of h2so4 is used to neutralize 0.3 m magnesium hydroxide solution, the volume of NaOH required to neutralize the given amount of Mg(OH)2 is 160 ml.

To answer this question, we need to use the concept of stoichiometry. Stoichiometry is the calculation of the quantities of reactants and products in a chemical reaction. The balanced chemical equation for the neutralization reaction between sulfuricacid (H2SO4) and magnesium hydroxide (Mg(OH)2) is:
H2SO4 + Mg(OH)2 → MgSO4 + 2H2O
From the equation, we can see that one mole of H2SO4 reacts with one mole of Mg(OH)2. Therefore, the number of moles of Mg(OH)2 in 60 ml of 0.3 m solution is:
moles of Mg(OH)2 = concentration x volume = 0.3 x (60/1000) = 0.018 moles
Since one mole of Mg(OH)2 requires one mole of NaOH to neutralize it, we need 0.018 moles of NaOH. The concentration of the NaOH solution is not given, so we cannot directly calculate the volume of NaOH required. However, we can use the concentration and volume of the H2SO4 solution to find the number of moles of H2SO4 used in the neutralization reaction:
moles of H2SO4 = concentration x volume = 0.40 x (60/1000) = 0.024 moles
From the balanced equation, we know that one mole of H2SO4 reacts with two moles of H2O. Therefore, the number of moles of H2O produced in the reaction is:
moles of H2O = 2 x moles of H2SO4 = 0.048 moles
Since the reaction is neutralization, the same number of moles of H2O and H+ ions are produced. Therefore, the number of moles of H+ ions produced is also 0.048 moles. Each mole of NaOH can neutralize one mole of H+ ions. Therefore, the number of moles of NaOH required is:
moles of NaOH = 0.048 moles
If we assume that the concentration of the NaOH solution is also 0.3 m, then the volume of NaOH required is:
volume of NaOH = moles of NaOH / concentration = 0.048 / 0.3 = 0.16 L = 160 ml

To learn more about stoichiometry, refer:-

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How many total atoms are in 0.380 g of P2O5

Answers

The gram formula mass of P2O5 is 31*2 + 16*5 =142. The mole number of P2O5 is 0.380/142=0.00268 mol. There is 7 atom in one P2O5 molecule. And there is 6.02 * 10^23 molecule per mol. So the answer is 1.13*10^22 atoms.

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