Consider the following reaction:2 HgCl2(aq) + C2O42-(aq) 2 Cl-(aq) + 2 CO2(g) + Hg2Cl2(s)(a) The rate law for this reaction is first order in HgCl2(aq) and second order in C2O42-(aq). What is the rate law for this reaction?Rate = k [HgCl2(aq)] [C2O42-(aq)]Rate = k [HgCl2(aq)]2 [C2O42-(aq)]Rate = k [HgCl2(aq)] [C2O42-(aq)]2Rate = k [HgCl2(aq)]2 [C2O42-(aq)]2Rate = k [HgCl2(aq)] [C2O42-(aq)]3Rate = k [HgCl2(aq)]4 [C2O42-(aq)](b) If the rate constant for this reaction at a certain temperature is 0.0169, what is the reaction rate when [HgCl2(aq)] = 0.158 M and [C2O42-(aq)] = 0.202 M?Rate = M/s.(c) What is the reaction rate when the concentration of HgCl2(aq) is doubled, to 0.316 M while the concentration of C2O42-(aq) is 0.202 M?Rate = M/s

Answer:

The pressure of the gas is kept constant. If the Kelvin temperature of the gas is doubled, the volume of the gas is. O 1.

Answer:

Molecular: [tex]Sr(NO_3)_2(aq) +Li_2SO_4(aq)\rightarrow SrSO_4(s)+2LiNO_3(aq)[/tex]

Ionic: [tex]Sr^{2+}(aq)+2NO_3^-(aq) +2Li^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)+2Li^+(aq)+2NO_3^-(aq)[/tex]Net ionic: [tex]Sr^{2}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)[/tex]

Explanation:

Hello!

In this case, since the molecular, ionic and net ionic equations show the complete molecules, ions and resulting ions respectively, for the reaction between strontium nitrate and lithium sulfate, we can notice the formation of solid strontium sulfate and lithium nitrate as shown below:

[tex]Sr(NO_3)_2(aq) +Li_2SO_4(aq)\rightarrow SrSO_4(s)+2LiNO_3(aq)[/tex]

Which is the molecular equation showing both reactants and products as molecules. Then, the ionic equation shows all the reactants and products as ions, considering that aqueous solutions dissociate whereas solid, liquid and gaseous molecules do not, therefore, we obtain:

[tex]Sr^{2+}(aq)+2NO_3^-(aq) +2Li^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)+2Li^+(aq)+2NO_3^-(aq)[/tex]

Finally, for the net ionic equation, we cancel out the spectator ions, which are those at both reactants and products:

[tex]Sr^{2+}(aq)+SO_4(aq)^{2-}\rightarrow SrSO_4(s)[/tex]

Best regards!

A further explanation is below.

Given:

Strontium nitrate reacts with Lithium sulfate just to produce Strontium sulfate ([tex]Sr(NO_3)_2[/tex]) and Lithium nitrate ([tex]Li NO_3[/tex]).

The molecular equation will be:

→ [tex]Sr(NO_3)_2(aq) +Li_2SO_4(aq) \rightarrow SrSO_4 (s) +2LiNO_3 (aq)[/tex]

The complete ionic equation will be:

→ [tex]Sr^{2+} (aq) +2NO_3^- (aq) +2Li^+ (aq)+ SO_4^{2-} (aq) \rightarrow SrSO_4 (s)+2Li^+ (aq) +2NO_3^- (aq)[/tex]

By removing the uncharged ions from equation's will be:

Spectator ion:

→ [tex]2Li^+ (aq), 2NO_3^- (aq)[/tex]

Net ionic equation will be:

→ [tex]Sr^{2+}(aq)+SO_4^{2-} (aq) \rightarrow SrSO_4 (s)[/tex]

Thus the response above is right.

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Answer:

3-methylcyclohexene

Explanation:

In gas chromatography, the compound with the lower boiling point will have shorter time in the gas phase, resulting in fewer theoretical plates and a decreased retention time. Once a product exits the column it is transferred to the graph after passing through a detector which picks up on the differing conductance of the carrier gas vs. the carrier gas and product

There are two types of gas chromatography, they are gas liquid chromatography and gas solid chromatography. Here 3 - methyl cyclohexene will elutes first as compared to 1-methylcyclohexene.

In gas liquid chromatography, the mobile phase is a gas and the stationary phase is a liquid immobilized on the surface of an inert solid support. The technique is based on the differential partitioning of the components of a sample in the vapor phase between the mobile gas phase and the stationary liquid phase.

In gas chromatography, the component with a lowest boiling point is found to has a shorter time in the gas phase and it will elute first as compared to the component with highest boiling point.

Here 3-methylcyclohexene is more substituted than 1-methylcyclohexene. The boiling point of 3-methylcyclohexene is less as compared to that of 1-methylcyclohexene. So 3-methylcyclohexene elute first.

Thus 3-methylcyclohexene elute first.

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Answer:

the density of CO2 gas at STP is 1.96 g/l.

Explanation:

see the pic for the answer

Answer:

hunting for other animals or when they are really hungry

Explanation:

We have that for the Question "Give an  example of when a plant or animal might  use energy they have stored. " it can be said that

From the question we are told

Give an  example of when a plant or animal might  use energy they have stored.

Generally

Plants may use the energy in food synthesizing and the may also use the stored energy in excretion.

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Answer:

B

Explanation:

a chemical change is making a new substance and physical change with a result in a different state of matter:)

Answer:

[tex]E=4.3\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The wavelength of a photon of blue light is 462 nm

We need to find the energy of this photon. The formula for the energy of a photon is given by :

[tex]E=\dfrac{hc}{\lambda}\\\\E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{462\times 10^{-9}}\\\\=4.3\times 10^{-19}\ J[/tex]

So, the energy of this photon is [tex]4.3\times 10^{-19}\ J[/tex].

Answer:

Boron

Explanation:

Because it has a complete 2s orbital and therefore, an increased shielding of the 2s orbital will reduce the ionisation energy.

Answer:

A group of cells doing the same job forms a tissue. A group of tissues working together forms an organ. Organs work in groups, too. A group of organs doing the same job is called a system.

Explanation:

Answer:

Following are the complete balance of the given equation:

Explanation:

Given equation:

[tex]Cu_2SO_4(aq) + Li_3PO_4(aq)[/tex]

[tex]Cu_2So4\ (aq)+Li_3Po_4 \(aq) \longrightarrow Cu_3(Po_4)\ (aq)+Li_2So_4 \ (aq)[/tex]

After Balancing the equation:

[tex]3 Cu_2So4\ (aq)+ 2 Li_3Po_4\ (aq) \longrightarrow 2 Cu_3(Po_4)\ (aq)+ 3Li_2So_4 \ (aq)[/tex]

In the above equation, when the 3 mol Copper sulfate reacts with 2 mol lithium phosphate , it will produce 2 mol Copper phosphate  and 3 mol Lithium sulfate .

Answer:

[tex]m_{K_2O}^{by\ K} =3.01gK_2O[/tex]

Explanation:

Hello!

In this case, since 2.50 g of both potassium (molar mass = 39.1 g/mol) and gaseous oxygen (molar mass = 32.0 g/mol) react in a 4:1 and 1:2 mole ratio respectively, to produce potassium oxide (molar mass = 94.2 g/mol), we evaluate the mass of potassium oxide yielded by each reactant in order to identify the limiting one via stoichiometry:

[tex]m_{K_2O}^{by\ K}=2.50gK*\frac{1molK}{39.1gK}*\frac{2molK_2O}{4molK}*\frac{94.2gK_2O}{1molK_2O} =3.01gK_2O\\\\m_{K_2O}^{by\ O_2}=2.50gO_2*\frac{1molO_2}{32.0gO_2}*\frac{2molK_2O}{1molO_2}*\frac{94.2gK_2O}{1molK_2O} =14.7gK_2O[/tex]

Thus, since the 2.50 g of potassium yields 3.01 g of potassium oxide, we infer it is the limiting reactant and that is the mass of produced product by the reaction.

Best regards!

Answer: Calculating mass quick check

1. The mass of the reactants must be equal to the mass of the products. The total number of moles of the reactants can be more or less than the total number of moles of the products.

2. Divide the mass of the reactant by its molar mass to find the number of moles of the reactant. Use the chemical equation to find the number of moles of the product. Multiply the number of moles of the product by its molar mass to find the mass of the product.

3. 2(108 g/mol)+32 g/mol=248 g/mol; (248 g/mol)(0.02 mol)=4.96 g

4. 19.5 g

5. 853.5 g

You're welcome

Answer:

309 K

Explanation:

Step 1: Convert the pressure to atm

We will use the conversion factor 1 atm = 760 mmHg.

732 mmHg × 1 atm/760 mmHg = 0.963 atm

Step 2: Calculate the moles (n) of the ideal gas

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.963 atm × 8.4 L/0.0821 atm.L/mol.K × 298 K

n = 0.33 mol

Step 3: Calculate the temperature change

We will use the following expression.

Q = n × Cp × ΔT

ΔT = Q/n × Cp

ΔT = 75 J/0.33 mol × 20.79 J/mol.K

ΔT = 11 K

Step 4: Calculate the final temperature

T = 298 K + 11 K = 309 K

Answer:

12.69g

Explanation:

as you see in the equation the number of moles for zinc hydroxide is same the water so,

0.705=mass\18mol g^-1

mass of water    = 12.69g

hope is helpful

Answer:

C. Alkaline Metal ok thanks

The energy of a photon that would move an electron from level 2 to level 5 : B. 4.8 eV

The electron energy at the nth shell can be formulated:

Rh = constant 2.179.10⁻¹⁸ J

So the electron transfer energy (ΔE)

ΔE = E final- E initial

energy at n=2(level 2) = -5.3 eV

energy at n=5(level 5) = -0.5 eV

So the energy absorbed :

[tex]\tt \Delta E=-0.5-(-5.3)=-0.5+5.3=4.8`eV[/tex]

The energy required to move an electron from level 2 to level 5 is 4.8 eV.

According to the Bohr model of the atom, an electron moves from a lower to a higher energy level when it absorbs energy. The same energy is released when the atom returns to ground state.

When an electron from level 2 to level 5, the energy of the photon required is;

ΔE = E5 - E2 = -0.5eV - (-5.3eV)

ΔE = 4.8eV

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Answer:

mineral crystal growth

acid rain

carbonic acid

Answer:

Mineral crystal growth, Acid rain, Carbonic acid

Answer:

Fraction of association = 0.01

pH = 11

Explanation:

A weak base, B, is in equilibrium with water as follows:

B(aq) + H2O(l) ⇄ BH⁺(aq) + OH⁻(aq)

Where Kb is defined as:

Kb = 1.00x10⁻⁵ = [BH⁺] [OH⁻] / [B]

Some B will react producing BH⁺ and OH⁻:

[BH⁺] = X

[OH⁻] = X

[B] = 0.100M - X

As Kb <<< [B] we can say:

[B] ≈ 0.100M

Replacing:

1.00x10⁻⁵ = [X] [X] / [0.100]

1.00x10⁻⁶ = X²

X = 1x10⁻³M = [BH⁺] = [OH⁻]

The fraction of association is [BH⁺] / [B] = 1x10⁻³M / 0.100M = 0.01

As [OH⁻] = 1x10⁻³M, pOH = -log[OH⁻] = 3

pH = 14- pOH

dicing potatoes, because the molecules are separated but remain the same substance

Answer:

For the people wondering, yes, dicing potatoes is the correct answer. Because a new substance has not formed.

Answer:

Like Roman said Avogadro's number is the key to your problem. The value that you'll use on paper is 6.022x10^23. Luckily in our case a mol specifically refers to the amount of atoms. Its a weird concept but 1.97 mol of Aluminum is the exact same amount of atoms as say helium. The only differences lay in the total mass of the sample.

Answer:

the answer is "Transferring of electrons from one atom to another creates the bond"

Explanation:

The statement which is true about covalent bonds is that the valence

electrons are shared in order to achieve the bond.

Covalent bonds involves the atoms of two elements sharing electrons in

order to achieve a stable configuration. The electrons which are shared are

those at the outermost layer of the shell and they are called valence

electrons.

These bonds help in the formation of new compounds such as water which

is formed from the covalent bonding between hydrogen and oxygen.

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Answer:

a. 167 mL b. 39.27 %

Explanation:

a. From the chemical equation. 2 mole of Al reacts with 3 mole H₂SO₄ to produce 1 mol  Al₂(SO₄)₃.

Now, we calculate the number of moles of Al in 45.0 g Al.

We know number of moles, n = m/M where m = mass of Al = 45.0 g and M = molar mass of Al = 26.98 g/mol.

So n = 45.0 g/26.98 g/mol = 1.668 mol

Since 2 mole of Al reacts with 3 mole H₂SO₄, then 1.668 mole of Al reacts with x mole H₂SO₄. So, x = 3 × 1.668/2 mol = 2.5 mol

So, we have 2.5 mol H₂SO₄.

Now number of moles of H₂SO₄, n = CV where C = concentration of H₂SO₄ = 15.0 M = 15.0 mol/L and V = volume of H₂SO₄.

V = n/C

= 2.5 mol/15.0 mol/L

= 0.167 L

= 167 mL of 15.0 M H₂SO₄ reacts with 45.0 g Al to produce aluminum sulfate.

b. From the chemical reaction, 2 mol Al produces 1 mol Al₂(SO₄)₃

Therefore 1.668 mol Al will produce x mol  Al₂(SO₄)₃. So, x = 1 mol × 1.668 mol/2 mol = 0.834 mol

So, we need to find the mass of 0.834 mol  Al₂(SO₄)₃. Now molar mass  Al₂(SO₄)₃ = 2 × 26.98 g/mol + 3 × 32 g/mol + 4 × 3 × 16 g/mol = 53.96 g/mol + 96 g/mol + 192 g/mol = 341.96 g/mol.

Also number of moles of  Al₂(SO₄)₃, n = mass of  Al₂(SO₄)₃,m/molar mass  Al₂(SO₄)₃, M

n =m/M

So, m = nM = 0.834 mol × 341.96 g/mol = 285.2 g

% yield = Actual yield/theoretical yield × 100 %

Actual yield = 112 g, /theoretical yield = 285.2 g

So, % yield = 112 g/285.2 g × 100 %

= 0.3927 × 100 %

=  39.27 %

The volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum is 166mL and % yield of the reaction is 39.46%.

Moles of any substance will be calculated by using the below formula as:
n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is :

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

Moles of 45g of Al will be calculated as:

n = 45g / 27g/mol = 1.66 mole

From the stoichiometry of the reaction, it is clear that:

1.66 moles of Al = react with 3/2×1.66=2.49 moles of H₂SO₄

By using the formula of molarity we can calculate the volume of H₂SO₄ as:

M = n/V

V = (2.49) / (15) = 0.166L = 166mL

Again from the stoichiometry it is clear that:
1.66 moles of Al = produces 1/2×1.66= 0.83 moles of Al₂(SO₄)₃

Mass of 0.83 moles of Al₂(SO₄)₃ = (0.83mol)(341.96g/mol) = 283.82 g

Given actual yield of Al₂(SO₄)₃ = 112g

% yield will be calculated as:

Percent yield = (Actual yield/Theoretical yield) × 100

% yield = (112/283.82) × 100 = 39.46%

Hence required values are discussed above.

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Answer:

Noble gases... (You already know...)

Explanation:

The right-most column of the periodic table contains atoms with full electron shells, also known as Noble Gases.

Hope this helped!

Answer:

Noble gases

Explanation:

The six elements listed in the last column of the table (Group 18) of the periodic table are collectively referred to as the noble gases. All of the noble gases have the maximum number of electrons possible in their outer shells (two electrons in helium’s outer shell, and eight electrons for all the others), making them stable.

Answer:

[tex]n_{H_2O}=3.0molH_2O\\\\n_{N_2}=1.0molN_2[/tex]

Explanation:

Hello!

In this case, for the described reaction we have:

[tex]2NH_3(g)+\frac{3}{2} O_2(g)\rightarrow N_2(g)+3H_2O(g)[/tex]

Which means there is 2:3/2 mole ratio between ammonia and oxygen and we use it to compute the consumed moles of ammonia by 13.0 moles of oxygen as shown below:

[tex]n_{NH_3}^{consumed \ by\ O_2}=13.0molO_2*\frac{2molNH_3}{\frac{3}{2}molO_2 } =17.33molNH_3[/tex]

However, since just 2.0 mol of ammonia is available, we infer it is the limiting reactant and the maximum amount of both nitrogen and water that can be produced is computed below:

[tex]n_{H_2O}=2.0molNH_3*\frac{3molH_2O}{2molNH_3} =3.0molH_2O\\\\n_{N_2}=2.0molNH_3*\frac{1molN_2}{2molNH_3} =1.0molN_2[/tex]

Best regards!