When 2.16g of H2 reacts with excess O2 by the following equation, 258 kJ of heat are released. What is the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas? 2H2+O2⟶2H2O

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

We have

[tex]V_2 = \frac{300 \times 2}{7} = \frac{600}{7} \\ = 85.714285...[/tex]

We have the final answer as

Hope this helps you

Answer:

CHO

Explanation:

Carbon = 41%,  Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12,         4.58/1,         54.6/16

3.41              4.58            3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41,      4.58/3.41,     3.41/3.41

1,                    1,                  1

Hence the empirical formula is CHO

Answer:

The empirical formula of X is C3H4O3.

Explanation:

Answer:

2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l )- combustion reaction

N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l )- decomposition reaction

C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) - combination reaction

2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g )- Redox reaction

C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s )- double displacement reaction

Explanation:

We can determine the type of reaction by considering the reactants and products.

Combustion is a reaction between a substance and oxygen which produces heat and light. The first reaction is the equation for the combustion of ethyne.

A decomposition reaction is one in which a single reactant breaks down to form products. The second reaction is the decomposition of ammonium nitrate.

A  combination reaction is said to occur when two elements or compounds react to form a single product. The third reaction is the combination of carbon dioxide and methane to form  methanol.

An oxidation-reduction reaction is a reaction in which there is a change in oxidation number of species from left to right of the chemical reaction equation. The fourth reaction is the oxidation of iron (0 to +3 state) and reduction of hydrogen (+1 to 0 state).

A double displacement reaction is a reaction in which ions exchange partners from left to right in the reaction equation. The fifth reaction is a double displacement reaction. Both Na^+ and Ca^2+ exchanged partners from left to right of the reaction equation.

Reactions are the formation of the products from the reactant. The types of reactions are combustion, decomposition, combination, Redox and double displacement.

The reaction is a chemical change in the properties of the reactant that forms the products. It can be of various types based on the formation of the product.

The first reaction is combustion as the reactants react and use oxygen to form heat, carbon dioxide and water. The combustion reaction of ethyne can be shown as,

[tex]\rm 2 C _{2} H _{2} ( g ) + 5 O _{2} ( g ) \rightarrow 4 C O _{2} ( g ) + 2 H _{2} O ( l )[/tex]

The second reaction is decomposition in which a single reactant decomposes to form two or more products. The decomposition of ammonium nitrate can be shown as,

[tex]\rm N H _{4} N O _{3} ( s ) \rightarrow N _{2} O ( g ) + 2 H _{2} O ( l )[/tex]

The third reaction is a combination reaction in which two compound or elements combines to form one product. The combination reaction between carbon monoxide and hydrogen to form methanol can be shown as,

[tex]\rm C O ( g ) + 2 H _{2} ( g ) \rightarrow C H _{3} O H ( l )[/tex]

The fourth reaction is redox and includes the oxidation and the reduction of the species of the reaction. In the reaction, iron undergoes oxidation and hydrogen reduction. The redox reaction can be shown as,

[tex]\rm 2 F e ( s ) + 6 H C l ( a q ) \rightarrow 2 F e C l _{3} ( a q ) + 3 H _{2} ( g )[/tex]

The fifth reaction is a double displacement reaction in which the calcium and sodium interchange their position in the product formation. The reaction can be shown as,

[tex]\rm C a C l _{2} ( a q ) + N a _{2} C O _{3} ( a q ) \rightarrow 2 N a C l ( a q ) + C a C O _{3} ( s )[/tex]

Therefore, the type of reactions is 1. combustion, 2. decomposition, 3. combination, 4. redox and 5. double displacement.

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Answer:

Exothermic

Explanation:

Nuclear fission means splitting, so there is a lot of energy being released.

Answer: The mass of blue copper sulfate is 3.5 g

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the heating of copper sulfate crystals is:

Let the mass of blue copper sulfate be 'x' grams

We are given:

Mass of copper sulfate powder = 2.1 grams

Mass of water = 1.4 grams

Total mass on reactant side = x

Total mass on product side = (2.1 + 1.4) g

So, by applying law of conservation of mass, we get:

Hence, the mass of blue copper sulfate is 3.5 grams

Answer:

Explanation:

NaNO₃  = Na⁺  + NO₃⁻¹

.497 M                 .497 M

moles of NO₃⁻¹ = .897 x .497 = .4458 moles

Ca( NO₃)₂  = Ca + 2 NO₃⁻¹

.341 M                    2 x .341 M = .682 M

moles of NO₃⁻¹ = .813 x .682 = .5544 moles

Total moles =  .4458 moles  +  .5544 moles

= 1.0002 moles

volume of solution = 897 + 813 = 1710 mL

= 1.710 L

concentration of nitrate ion = 1.0002 / 1.710 M

= .585 M

Answer:

the second answer its science behind it

Answer:

b

Explanation:

Answer:

4.22mL

Explanation:

V=m/d

v= 18.45g/4.37g/mL

Complete Question

You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the enantiomeric excess?

Answer:

The value is  [tex]k  = 96 %[/tex]

Explanation:

From the question we are told that

   The  percentage of  enantiomer A is  A =  98%

    The  percentage of  enantiomer B is  B =  2%

Generally the enantiomeric excess is mathematically represented as

      [tex]k  =  \frac{A -B}{A+B}  *  100[/tex]

=>   [tex]k  =  \frac{98 -2}{98+2}  *  100[/tex]

=>   [tex]k  = 96 %[/tex]

Answer:

a. 0.183 mol C

b. 0.366 mol H

Explanation:

Assuming total combustion, all of the carbon in the unknown compound turned into carbon dioxide, CO₂.

So first we calculate the CO₂ moles produced, using its molecular weight:

This means in the unknown compound there were 0.183 moles of carbon, C.

Conversely, all of the hydrogen in the unknown compound turned into water, H₂O.

Calculating the H₂O moles:

We multiply the water moles by two, as there are 2 H moles per H₂O mol:

Answer:

I think it's A and D

Explanation:

I'm not sure if it's right

Answer:

The answer is B and D

Explanation:

trust fr

Answer:

2,2,3- trimethylhexane because it has more carbon atoms than pentane.

Explanation:

Hello.

In this case, since the physical properties of organic compounds are intensified as the number of carbon atoms start increasing on it, the larger the amount of carbon atom, the higher the boiling point since more energy is required to allow the liquid-phase molecules to transcend to the vapor-phase.

In such a way, since pentane has five carbon atoms and 2,2,3- trimethylhexane has nine carbon atoms, 2,2,3- trimethylhexane has the highest boiling point.

Best regards.

C – 19,9%, H – 2,2%, N – 11,6%, O – x%

[tex]M=120\frac{g}{mol}[/tex]

1 percentage

The entire molecule is 100% and all the components of the compound add up to 100%.

100% - 19,9% - 2,5% - 11,6% = 66,1%

The compound contains 66,1% oxygen.

2 molar masses

[tex]M_{C}=12,01\frac{g}{mol}[/tex]

[tex]M_{H}=1,008\frac{g}{mol}[/tex]

[tex]M_{O}=15,999\frac{g}{mol}[/tex]

[tex]M_{N}=14,007\frac{g}{mol}[/tex]

3 masses

The compound has a molar mass of 120g/mol. So one molecule weighs 120 g. To find out how much the percentage of a component weighs, you have to calculate it using the molar mass.

carbon

19,8% of 120g

[tex]m=120g*0,198\\m=23,76g[/tex]

One molecule contains 23,76g of carbon.

hydrogen

2,5% of 120g

[tex]m=120g*0,025\\m=3g[/tex]

One molecule contains 3g of hydrogen.

oxygen

66,1% of 120g

[tex]m=120g*0,661\\m=79,32g[/tex]

One molecule contains 79,32g of oxygen.

nitrogen

11,6% of 120g

[tex]m=120g*0,0,116\\m=13,92g[/tex]

One molecule contains 13,92g of nitrogen.

4 amount of substance

carbon

[tex]n=\frac{23,76g}{12,01\frac{g}{mol} }\\n=1,98mol[/tex]

The compound contains about 2 moles of carbon.

hydrogen

[tex]n=\frac{3g}{1\frac{g}{mol} }\\n=3mol[/tex]

The compound contains about 3 moles of hydrogen.

oxygen

[tex]n=\frac{79,32g}{15,999\frac{g}{mol} }\\n=4,96mol[/tex]

The compound contains about 5 moles of oxygen.

nitrogen

[tex]n=\frac{13,92g}{14,007\frac{g}{mol} }\\n=0,99mol[/tex]

The compound contains about 1 moles of nitrogen.

5. molecular formula

The formula results from the ratio of the amounts of substance.

[tex]n_{C} :n_{H} :n_{O} :n_{N} =2:3:5:1\\C_{2}H_{3}NO_{5}[/tex]

The molecular formula of the given compound is C₂H₃NO₅, and percent composition of oxygen in it is 66.1%.

Mass of any composition of any compound will be calculated by using the below formula as:

Mass of component = (% composition)×(mass of compound) / 100

Given mass of compound = 120g/mol

Total composition of compound (100%) = Percent composition of all components

% composition of oxygen = 100 - (19.8 + 2.50 + 11.6) = 66.1%

Moles will be calculated as:

n = W/M, where

W = given mass

M = molar mass

Mass of Carbon component = (0.198)(120g) = 23.76g

Moles of Carbon atom = 23.76g / 12.01g/mol = 1.98mol = 2 moles

Mass of Nitrogen component = (0.116)(120g) = 13.92g

Moles of Nitrogen atom = 13.92g / 14.007g/mol = 0.99mol = 1 moles

Mass of Oxygen component = (0.661)(120g) = 79.32g

Moles of Oxygen atom = 79.32g / 15.99g/mol = 4.96mol = 5 moles

Mass of Hydrogen component = (0.025)(120g) = 3g

Moles of Hydrogen atom = 3g / 1g/mol = 3 moles

So, the molecular formula of the compound on the basis of moles of given entities is C₂H₃NO₅.

Hence required molecular formula is C₂H₃NO₅.

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Answer:

5;11

Explanation:

Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons

Answer:

A) enthalpy of reaction

Explanation:

The region C signifies the enthalpy of reaction.

This diagram is the energy profile of an endothermic reaction. In such reaction, heat is absorbed from the surrounding. At the end of the reaction, the heat of product is lesser than that of the reactants.

it is indeed A) enthalpy of reaction

Answer:

Explanation:

Iodine is the most flexible of the atoms in Group 17 of the periodic table. It can have more than 1 charge, especially when combined with oxygen. The answer that you are expected to give, I think is C. Iodine's dominant charge is - 1

Answer:

Explanation:

Al(s) + 3/2 Br₂(l) = AlBr₃(s)

K = [  AlBr₃] / [ Al] [  Br₂]³/²

K² =  [  AlBr₃]² / [  Al ] ² [ Br₂]³

2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =

K₁ =  [  Al ] ² [ Br₂]³ /  [  AlBr₃]²

K₁ =  ( 1 / K² ) = K⁻²

2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)

K₂ = [ AlBr₃ ]² / [  Al ]² [  Br₂ ]³

K₂ = K²

3 )

AlBr₃(s) =   Al(s) + 3/2 Br₂(l)

K₃  = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]

=  ( 1 / K ) = K⁻¹

Calcium (Ca) is in group 2 and period 4 on the periodic table be because Calcium has 2 valence electrons and 4 electron shell. Thus, calcium is a metal like all other group 2 element.

The correct answer to the question is Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.

Calcium is a group 2 element majorly because it has 2 valence electrons. It is also in period 4 because it has 4 electron shells.

Being a group 2 element, calcium is a solid at room temperature and also reactive. All elements in the group 2 are metals.

There are other elements in period 4 which are not solid. For example krypton is an element in period 4 and it is a gas and not reactive.

From the above information, we can conclude that the correct answer to the question is:

Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.

See attached image

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Answer:

5Ns

momentum= mass *velocity

=1*5

=5Ns

Answer:

2.35 x 10²⁰ atoms Ga

Explanation:

After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.

[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]

Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.

The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms

From the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.

First, we will determine the number of moles of Ga present

Using the formula,

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]

Mass = 27.2 mg = 0.0272 g

Molar mass = 69.72 g/mol

Then,

[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]

[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles

Now, for the number of atoms present

From the formula

Number of atoms = Number of moles × Avogadro's constant

Then,

Number of Ga atoms = 0.000390132 × 6.022×10²³

Number of Ga atoms = 2.35 × 10²⁰ atoms

Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms

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Answer:

Mass = 23.232 g

Explanation:

Given data:

Mass of C₈H₁₈ = 7.58 g

Mass of CO₂ produced = ?

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂      →     16CO₂ + 18H₂O

Number of moles of octane:

Number of  moles =  mass/molar mass

Number of moles = 7.58 g/ 114.23 g/mol

Number of moles = 0.066 mol

Now we will compare the moles of CO₂ with octane from balance chemical equation.

                  C₈H₁₈            :           CO₂

                       2              :              16

                 0.066             :           16/2×0.066 = 0.528

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 0.528 mol × 44 g/mol

Mass = 23.232 g

Answer:

polar bonds are caused by different kind of atoms, because almost every atoms have different powers to attract electrons.

the answer will be the two same atoms, F2

Answer:

The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].

The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].

Explanation:

The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.

The Rydberg Equation gives the following relation:

[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].

Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:

[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].

In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:

[tex]\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}[/tex].

Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].

Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:

[tex]\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}[/tex].

Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].

Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:

[tex]\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}[/tex].

The energy of the transition between n = 7 and n = 3  is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.

Using the Rydberg Equation for energy;

ΔE = -RH(1/n^2final - 1/n^2initial)

Given that;

nfinal = 3

ninitial = 7

RH = 2.18 × 10^-18 J

ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)

ΔE = - 2.18 × 10^-18(0.11 - 0.02)

ΔE = - 1.96 × 10^-19 J

For the second part;

Since the wavelength is 1000nm, we have;

λ = 1000nm

c = 3 × 10^8 m/s

f = ?

c = λf

f = c/λ

f = 3 × 10^8 m/s/1000 × 10^-9 m

f = 3 × 10^8 m/s/ 1 × 10^-6 m

f = 3 × 10^14 Hz

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Answer:

Kc = 4.86×10⁻⁶

Explanation:

We begin from the equation:

N₂  +  2H₂ ⇄  2NH₃

We start from 3.47×10⁻² moles of N₂(g) and 6.38×10⁻² moles of H₂(g), so when we reach the equilibrium, we get 6.25×10⁻² moles of H₂.

This data means, that in the reaction we made react:

6.38×10⁻² - x = 6.25×10⁻²

x = 1.3×10⁻³ moles of H₂

As stoichiometry is 1:3, we will know that the moles of N₂ that have been reacted were:

1.3×10⁻³ moles / 3 = 4.33×10⁻⁴ moles of N₂

So, in the equilibrium we would have:

3.47×10⁻² moles of N₂ - 4.33×10⁻⁴ moles of N₂ = 0.0343 moles of N₂

How many ammonia, would we have in the equilibrium?

4.33×10⁻⁴ mol . 2 = 8.66×10⁻⁴ moles (from stoichiometry with N₂, 1:2)

(1.3×10⁻³ mol . 2) / 3 = 8.66×10⁻⁴ moles (from stoichiometry with H₂, 2:3)

Let's make the expression for Kc

Kc = [NH₃]³ / [N₂] . [H₂]²

(8.66×10⁻⁴ )³ / (0.0343 . (6.25×10⁻²)² = 4.86×10⁻⁶