The equation for the reaction of magnesium ribbon and hydrochloric acid is
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
If 1.53 g magnesium reacted, how many moles of hydrogen gas was produced?
Report your answer to three significant figures.
Mg = 24.3 g/mol

Answer:

Sphere

Explanation:

The shape of the s-sublevel is spherical in shape.

This sublevel has an azimuthal quantum number of 0 and it is spherical in shape;

Answer:

it is C

Explanation:

because Potassium iodide (KI) is an ionic compound which is made of the following ions: K+I−

Answer:

The correct answer would be C.) a line graph

Explanation:

I just did the assignment on edge 2020 and got it right

Answer:

ether group

Explanation: I looked it up

Answer:

See Explanation

Explanation:

Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.

This implies that this compound Pb2O3  (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.

Hence Pb2O3  contains PbO and PbO2 units.

Answer:

[tex]23.093\ \text{torr}[/tex]

Explanation:

[tex]M_g[/tex] = Molar mass of glucose = 180.2 g/mol

[tex]M_w[/tex] = Molar mass of water = 18 g/mol

[tex]m_g[/tex] = Mass of glucose = 76.6 g

[tex]m_w[/tex] = Mass of water = [tex]250\times 1\ \text{g/mL}=250\text{g}[/tex]

[tex]P_0[/tex] = Vapor pressure of pure water at 25°C = 23.8 torr

The mole fraction of glucose is

[tex]x_g=\dfrac{\dfrac{m_g}{M_g}}{\dfrac{m_g}{M_g}+\dfrac{m_w}{M_w}}=\dfrac{\dfrac{76.6}{180.2}}{\dfrac{76.6}{180.2}+\dfrac{250}{18}}\\\Rightarrow x_g=0.0297[/tex]

Mole fraction of the solute would be

[tex]\dfrac{P_0-P}{P_0}=x_g\\\Rightarrow 0.0297=\dfrac{23.8-P}{23.8}\\\Rightarrow P=23.8-0.0297\times23.8\\\Rightarrow P=23.093\ \text{torr}[/tex]

The vapor pressure of the solution is [tex]23.093\ \text{torr}[/tex].

Answer:

there is your answer hope it helped :3

Answer:

[tex]k=8.63x10^{-8}\frac{1}{M*s}[/tex]

Explanation:

Hello!

In this case, since the differential rate law of a second-order reaction is:

[tex]\frac{dC_A}{dt}=-kC_A^2[/tex]

Whereas A stands for NOCl and the corresponding integrated rate law is:

[tex]\frac{1}{C_A} =kt+\frac{1}{C_A_0}[/tex]

Thus, since we know the concentrations and the elapsed time, we compute the rate constant as shown below:

[tex]k=( \frac{1}{C_A}-\frac{1}{C_A_0} )/t\\\\k=( \frac{1}{0.83M}-\frac{1}{0.878M} )/763,200s\\\\k=8.63x10^{-8}\frac{1}{M*s}[/tex]

Best regards!

Answer:

1.5 moles of Fe produced.

Explanation:

Given data:

Moles of FeO react = 1.50 mol

Moles of iron produced = ?

Solution:

Chemical equation:

FeO + CO       →       Fe + CO₂

Now we will compare the moles of ironoxide with iron.

                           FeO          :           Fe

                              1             :             1

                             1.5           :           1.5

Thus from 1.5 moles of FeO 1.5 moles of Fe are produced.                                    

1.5 moles of Fe can be obtained when 1.50 mol of FeO reacts with an excess

of CO

Moles of FeO reacted  = 1.50 mol

Moles of iron produced = ? mol

The chemical equation is given as

FeO + CO       →       Fe + CO₂

Comparisons of FeO and Fe show they are in the ratio 1 : 1

                          FeO          :           Fe

                            1.5           :           1.5

which translates to the number of moles being equal

Therefore, 1.5 moles of FeO 1.5 moles of Fe are produced.    

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Answer:

Explanation:

25 Mass

The mass of the aluminum that has reacted is equal to 2.84 grams.

A chemical equation of a reaction contains of reactants, formed products, and an arrow indicating the direction of the given reaction. The balanced chemical equation in which the number of atoms is equal on either side of the equation. The law of conservation of mass should be followed by a balanced equation.

The chemical equation of the reaction of the KOH and Al is :

[tex]Al + KOH + H_2O \longrightarrow KAl(OH)_4 + H_2[/tex]

From the above equation, we can say that one mole of aluminum is reacting with one mole of KOH.

Given the volume and concentration of KOH are 35 ml and 3.0 M.

The moles of KOH = 3 × 0.035 = 0.105 mol

Therefore, the moles of Al will react = 0.105 mol

The mass of Al will react with KOH = 0.105 × 27 = 2.84 g

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Answer:

695.8kJ/mol = Enthalpy of combustion

Final temperature = 407.4°C

Explanation:

Taken the reaction:

2CH₃NO₂ + 3/2 O₂ → 2CO₂ + 3H₂O + N₂

The enthalpy of the reaction is the heat released when 1 mole of nitromethane are in combustion.

We need to find the moles presents in 121L of nitromethane using its density and molar mass (61.04g/mol). As these moles releases 1.6x10⁶kJ, we need to find heat released per mole of nitromethane

Mass nitromethane:

121L = 121000cm³ * (1.16g/cm³) = 140360g

Moles:

140360g * (1mol / 61.04g) = 2299.5 moles of nitromethane

Heat released per mole of nitromethane:

1.6x10⁶kJ / 2299.5 mol = 695.8kJ/mol = Enthalpy of combustion

The symbol - is because the heat is released, a + symbol is when the heat is absorbed.

Using:

Q = m×C×ΔT

Where Q is heat (1.6x10⁶kJ = 1.6x10⁹J)

m is mass of water:

1kL = 1000L = 1x10⁶mL = 1x10⁶g

C is specific heat of water: 4.184J/g°C

And ΔT is change in temperature.

Solving for ΔT:

1.6x10⁹J = 1x10⁶g * 4.184J/g°C * ΔT

ΔT = 382.4°C

As initial temperature was 25°C, final temperature is:

382.4°C + 25°C =

407.4°C

K20

K = 1+

0 = 2-

K/0 = 1/2

the answer is K:O = 1:2

Answer:

Answer: Tin (II) nitride

Answer:

maalat ang salt

Explanation:

Answer:

its B

Explanation:

Because its can only go cold to hot. -hope this help :)

Answer:

A photon

Explanation:

When electrons pass close to the standing wave of light, they hit the particles that are called photons

Answer: 10

Explanation:

4Al + 2O2 --> 2Al2O3

This is the balanced equation so there are 2*2 +2*3 = 4+6 =10

Answer:

E = 3.6×10⁻¹⁹ J

Explanation:

Given data:

Wavelength = 550 nm  (550 ×10⁻⁹ nm)

Energy of wave = ?

Solution:

Formula:

E = h c/λ

c = 3×10⁸ m/s

h = 6.63×10⁻³⁴ Js

Now we will put the values in formula.

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s /550 ×10⁻⁹ nm

E = 19.89×10⁻²⁶ J.m /550 ×10⁻⁹ nm

E = 0.036×10⁻¹⁷ J

E = 3.6×10⁻¹⁹ J

Answer:

Transition metals can have multiple oxidation states because of their electrons. The transition metals have several electrons with similar energies, so one or all of them can be removed, depending the circumstances. This results in different oxidation states.

Answer:

The answers are "partially filled" and "d" on Plato

Explanation:

Answer:

2HBr + O2 = H20 +Br2

Explanation:

Answer:

4hbr(aq) + O2(g) = 2H2O + 2Br2

Explanation:

Answer:

0.548 moles of HBr are required

Explanation:

Given data:

Number of moles of  hydrobromic acid = ?

Moles of bromine formed = 0.274 mol

Solution:

Chemical equation:

2HBr     →   H₂  + Br₂

Now we will compare the moles of HBr with Br₂.

               Br₂           :           HBr

               1               :            2

               0.274       :           2×0.274=0.548

Thus, 0.548 moles of HBr are required.

Answer:

[tex]T_b=239.7K=-33.49\°C[/tex]

Explanation:

Hello!

In this case, since the relationship between entropy and enthalpy for any process is defined below:

[tex]S=\frac{H}{T}[/tex]

For the vaporization of ammonia or any liquid, we can write:

[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T_{vap}}[/tex]

In such a way, solving the temperature of vaporization, or boiling point, we have:

[tex]T_{vap}=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]

Plugging in the given enthalpy and entropy of vaporization we obtain:

[tex]T_{vap}=T_b=\frac{23350\frac{J}{mol} }{97.43\frac{J}{mol*K}} \\\\T_b=239.7K=-33.49\°C[/tex]

Best regards!

From the information provided in the question, the boiling point of ammonia is 240 K.

Entropy is the degree of disorderliness in a system. The entropy of a system can be obtained using the relation;

ΔS°vap = ΔH°vap /T

Now;

ΔH°vap = 23.35 kJ/mol

ΔS°vap = 97.43 J/mol·K

T = ?

Making T the subject of the formula and substituting values;

T = ΔH°vap /ΔS°vap

T = 23.35 × 10^3J/mol/97.43 J/mol·K

T = 240 K

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Answer:

1.6 L CH3OH

Explanation:

3.20 L H2 x (1 mol H2/22.4 L H2) x (1 mol CH3OH/2 mol H2) x (22.4 L CH3OH/1 mol CH3OH) = 1.60 LCH3OH

1.602 liters of CH₃OH gas are formed when 3.20 L of H₂ gas are completely reacted at STP according to the following chemical reaction:

CO(g) + 2 H₂(g) → CH₃OH(g)

Remember 1 mol of an ideal gas has a volume of 22.4 L at STP

To find the volume of CH₃OH gas formed when 3.20 L of H₂ gas is completely reacted, we need to use stoichiometry and the given balanced chemical equation.

The balanced equation is:

CO(g) + 2 H₂(g) → CH₃OH(g)

From the balanced equation, we can see that 1 mol of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH.

Step 1: Convert the given volume of H₂ gas to moles at STP.

Using the information provided, 1 mol of any ideal gas at STP (Standard Temperature and Pressure) occupies 22.4 L.

Number of moles of H₂ gas = Volume of H2 gas at STP / Volume of 1 mol of H₂ gas at STP

Number of moles of H₂ gas = 3.20 L / 22.4 L/mol

Number of moles of H₂ gas = 0.14286 moles (rounded to 5 decimal places)

Step 2: Use stoichiometry to find the moles of CH₃OH formed.

From the balanced equation, we know that 1 mol of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH.

Since the ratio of H₂ to CH₃OH is 2:1, the number of moles of CH₃OH formed will be half of the moles of H₂ used in the reaction.

Number of moles of CH₃OH = (1/2) * Number of moles of H₂

Number of moles of CH₃OH = (1/2) * 0.14286

Number of moles of CH₃OH = 0.07143 moles (rounded to 5 decimal places)

Step 3: Convert the moles of CH₃OH to volume at STP.

Using the information provided, 1 mol of any ideal gas at STP (Standard Temperature and Pressure) occupies 22.4 L.

Volume of CH₃OH gas at STP = Number of moles of CH₃OH * Volume of 1 mol of CH₃OH gas at STP

Volume of CH₃OH gas at STP = 0.07143 moles * 22.4 L/mol

Volume of CH₃OH gas at STP = 1.602 L (rounded to 3 decimal places)

Therefore, approximately 1.602 liters of CH₃OH gas are formed when 3.20 liters of H₂ gas are completely reacted at STP.

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Answer:

I believe its 1,2, and 5

Explanation:

C D E

temperature of medium

type of wave

type of medium

Answer:

From the numerical steps highlighted under explanation, the average atomic mass of bromine is 79.91 u

Explanation:

The steps to be taken will involve;

1) Find the number of isotopes of bromine.

2) Identify the atomic mass and relative abundance of each of the isotopes.

3) Multiply the atomic mass of each of the isotopes by their corresponding values relative abundance value.

4) Add the value in step 3 above to get the average atomic mass of bromine.

Now;

Bromine has 2 isotopes namely;

Isotope 1: Atomic mass = 78.92amu and a relative abundance of 50.69%.

Isotope 2: Atomic mass = 80.92amu and a relative abundance of 49.31%.

Using step 3 above, we have;

(78.92 × 50.69%)

And (80.92 × 49.31%)

Using step 4 above, we have;

(78.92 × 50.69%) + (80.92 × 49.31%) ≈ 79.91 u

The atomic mass of bromine has been calculated as 79.9062 amu.

The relative atomic mass of the element has been given as the mass of each isotope with respect to their abundance.

The atomic mass has been given as:

[tex]amu=\sum mass\;\times\;\% Abundance[/tex]

The available isotopes of Bromine have been:

Isotope 1 = 78.92 amu, and 0.5069 % abundance

Isotope 2= 80.92 amu, and 0.4931 % abundance

Submitting the values for the atomic mass unit of bromine ([tex]amu_{\text{Br}}[/tex]):

[tex]amu_{\text {Br}}= (78.92\;\times\;0.5069)\;+\;(80.92\;\times\;0.4931)\\amu_{\text{Br}}=40.0045\;+\;39.9016\\amu_{\text{Br}}=79.9062[/tex]

The atomic mass of bromine has been calculated as 79.9062 amu.

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Answer:

the mole fraction of water in a solution is 0.17

Explanation:

The computation of the mole fraction of water in a solution is shown below:

Given that

Ethanol be 8.0 mol

And, the water be 1.6 mol

Based on the above information, the mole fraction of water in a solution is

= Water ÷ (Water + ethanol)

= 1.6 mol ÷ (1.6 mol + 8.0 mol)

= 1.6 mol ÷ 9.6 mol

= 0.17

Hence, the mole fraction of water in a solution is 0.17

Answer:

You didn't provide any statements. If it is an option, art is a form of expression. Something created from imagination that expresses the thoughts and emotions of the artist.

The statement that best answers the question: What is art is that—Art is a form of visual language. Therefore, the correct option is A.

Art is a diverse range of human activities and products that are created with the intention of expressing the imagination, emotions, or ideas, or to make a statement, in a visually appealing way.

Art can take many forms, including visual arts (such as painting, sculpture, and photography), performing arts (such as music, dance, and theater), and literary arts (such as poetry and prose).

It is typically seen as a form of communication and can be used to convey ideas, emotions, or political, social or cultural messages. It can also be used to simply evoke a response or emotion in the viewer, or to create a sense of beauty or aesthetic enjoyment. Therefore, the correct option is A.

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The question is incomplete, but most probably the complete question is,

Which of these statements best answers the question: What is art?

A. Art is a form of visual language.

B. Art is a form of written language.

C. Art is a form of spoken language.

D. Art is a form of dance language.

Answer:

(a) 6332.812 N (b) 1469.59  pounds.

Explanation:

Given that,

The surface area of the top of yu head is 25 cm x 25 cm

Area = 625 cm² = 0.0625 m²

Atmospheric pressure on it is 101325 Pa.

(a) Pressure acting on an object is given by force acting per unit area. It can be written as :

[tex]P=\dfrac{F}{A}\\\\F=PA\\\\F=101325\ Pa\times 0.0625\ m^2\\\\F=6332.812\ N[/tex]

6332.812 N of force is acting on your head at sea level.

(b) If area, A = 100 in²

We know that,

1 atm = 14.6959 pounds/in²

[tex]P=\dfrac{F}{A}\\\\F=PA\\\\F=14.6959 \ \text{Pounds}/in^2\times 100\ in^2\\\\F=1469.59\ \text{pounds}[/tex]

So, the force is 1469.59  pounds.

A. The force required to push your head at sea level is 6332.8125 N

B. The force in pound is 1469.59 lb

A. Determination of the force required at sea level.

Pressure (P) = 101325 Nm²

Area (A) = 25 cm × 25 cm = 0.25 m × 0.25 m = 0.0625 m²

P = F / A

Cross multiply

F = PA

F = 101325 × 0.0625

Therefore, the force required to push your head at sea level is 6332.8125 N

B. Determination of the force in pounds

Area (A) = 100 in²

Pressure (P) = 14.6959 pound / in²

F = PA

F = 14.6959 × 100

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