please help asap!

3. A double replacement reaction occurs between two solutions of lead (II) nitrate and potassium bromide. Write a
balanced equation for this reaction-identifying the product that will precipitate, and the product that will remain in
solution.
a) Write the balanced equation for this double replacement reaction.
b) If this reaction starts with 32.5 g lead (II) nitrate and 38.75 g potassium bromide, how many grams of the
precipitate will be produced? Remember to use the limiting reactant to calculate the amount of precipitate
formed.
c) How many grams of the excess reactant will remain?

Answer:

Density = 0.17 g/L

Explanation:

It is given that,

Volume of the inflated balloon filled with Helium, \(V=2.2\times 10^3\ L\)

Mass, m = 374 g

We need to find the density of helium. It is equal to its mass per unit volume. It can be given by :

d =m/V

\(d=\dfrac{374\ g}{2.2\times 10^3\ L}\\\\=0.17\ g/L\)

So, the density of helium in the balloon is 0.17 g/L.

The predominant form of ethylenediamine at pH 6.184 is H2NCH2CH2NH+3.

At pH 6.184, which is slightly acidic, the amino groups in ethylenediamine (H2NCH2CH2NH2) can partially protonate, resulting in the formation of ammonium ions (H+3NCH2CH2NH+3). The equilibrium between the neutral form and the protonated form is pH-dependent.

Therefore, at pH 6.184, the predominant form of ethylenediamine is H2NCH2CH2NH+3.

The final pressure of the gas mixtures is 72.48 atm.

The given parameters;

The number of moles of the gases is calculated as follows;

\(number \ of \ moles \ of \ oxygen = \frac{88}{16} = 5.5 \ moles\\\)

\(number \ of \ moles \ of \ argon = \frac{33}{40} = 0.825 \ moles\\\)

The total number of moles of the gas mixture;

= 5.5 + 0.825

= 6.325 moles

The final pressure of the mixture is calculated as follows;

PV = nRT

\(P = \frac{nRT}{V} \\\\P = \frac{6.325\ mol \times (0.0821 \ L.atm/mo.K) \times (62+273)}{2.4 \ L} \\\\P = 72.48 \ atm\)

Thus, the final pressure of the gas mixtures is 72.48 atm.

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Mn(s) metal will spontaneously react with Zn²⁺(aq), but will not

spontaneously react with Mg²⁺ (aq)

The Eo value of an electrochemical cell determines its spontaneity. Positive Eo electrochemical cells are spontaneous, and vice versa.

The relevant Eo of the half-cell in this instance are as follows for Mn(s) metal

Zn2+/Zn = -0.763v for Eo

2.37v for Eo Mg2+/Mg.

Mn2+/Mn = -1.18v for Eo.

Consequently, the equation for an Eo cell (with Zn as one of the half-cells) is: Eo Zn2+/Zn - Eo Mn2+/Mn = -0.763 - (-1.18) = 0.417v.

On the other hand, the equation for an Eo cell (with Mg as one of the half-cells) is: Eo Mg2+/Zn - Eo Mn2+/Mn = -2.37 - (-1.18) = -1.19v.

As a result, Mn(s) metal will spontaneously react with Zn2+(aq), but not with Mg2+ (aq)

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Answer:

heat wave

Explanation:

A high pressure system is a whirling mass of cool, dry air that generally brings fair weather and light winds. When viewed from above, winds spiral out of a high-pressure center in a clockwise rotation in the Northern Hemisphere.

An example of heterogeneous equilibrium is:

Heterogeneous equilibrium refers to an equilibrium state in a chemical reaction where the reactants and products exist in different physical states or phases. It occurs when substances in different phases, such as solids, liquids, and gases, are involved in a chemical reaction.

Considering the given equations:

The equation I: FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g) represents a heterogeneous equilibrium.

This is because the reactants and products involve different phases (solid and gas). FeO is a solid (s), CO is a gas (g), Fe is a solid (s), and CO₂ is a gas (g). The reaction involves the conversion of a solid and a gas to another solid and a gas, and the equilibrium is established between these different phases.

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Answer:

I think the answer is B :)

Answer:

the first step is B

Explanation: I'm on this lesson

Reaction of O₂ with HCl is as follow,

                                 4 HCl  +  O₂    →    2 H₂O  +  2 Cl₂

Let suppose we are given with 1 mole of HCl gas, Then,

According to eq,

  89.6 L (4 moles) of HCl when reacted produced  = 44.8 L (2 moles) Cl₂ gas

So,

                      22.4 L (1 mole) of HCl will produce  =  X L of Cl₂ gas

Solving for X,

                              X  =  (22.4 L × 44.8 L) ÷ 89.6 L

                              X  =  11.2 L of Cl₂ Gas

Fe(OH)₂(s) + O₂(g) +OH → Fe(OH)₃(s) + H₂O balance in basic solution is 4Fe(OH)₂ (s) + O₂(g) + 2H₂O → 4Fe(OH)₃(s)

A basic solution is an aqueous solution containing more OH⁻ ions than H⁺ ion

The unbalanced redox equation is as follows:

Fe(OH)₂(s) + O₂(g) +OH → Fe(OH)₃(s) + H₂O

In the chemical reaction all atoms other than hydrogen and oxygen are balanced and the oxidation number of fe changes from 2 to 3 and the change in the oxidation number is 1 and the oxidation number of fe changes from 0 to -2 and for 2O atoms the change in the oxidation number is 4 and then balance the equation with increase in the oxidation number and with decrease in the oxidation number

Fe(OH)₂(s) and Fe(OH)₃(s) with 4 then,

4Fe(OH)₂(s)+O₂(g) → 4Fe(OH)₃(s)

To balance O atoms, add 2 water molecules on LHS,

4Fe(OH)₂ (s) + O₂(g) + 2H₂O → 4Fe(OH)₃(s)

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Answer:

1600J

Explanation:

1 mole of oxygen gas (O2) has a mass of 32g i.e. the molar mass = 32g/mol

Kinetic energy (K.E) = ½ × m × v²

Where;

m = mass (g)

v = speed or velocity (m/s)

From the information given in this question;

m = 32g

V = 10m/s²

K.E = ½ × 32 × 10²

K.E = 16 × 100

K.E = 1600J

Answer:

D

Explanation:

Answer:

Explanation:

A )  (NH₄)₃PO₄(aq) + Bi(NO₃)₃(aq) =  BiPO₄(s) + 3 NH₄NO₃(aq).

B₁³⁺ +   PO₄⁻³  =  Bi ( PO₄ )

B )  AgOH(s) + H₂SO₄ (aq)  =  AgSO₄(s) + H₂O.

Ag⁺  + OH⁻ + H⁺ +  2 SO₄⁻² (aq)  =  Ag(SO₄)₂(s) + H₂O.

C )  HCIO₂ (aq) + Mn(OH)₂(s) =   Mn(Cl0₂)₂ (aq) + H₂O.

     H⁺  + OH⁻  =  H₂O .

Answer:

4 M

Explanation:

See the steps and explanation in the picture

Answer:

424.7J-312J=the specific amount of heat passing through the metal

b

b357

Answer:

2Fe +3 H2SO4  Fe2(SO4)3 + 3H2

2*56g. 3*98g➔400g. +6g

we have

2*56 g of Fe gives 400g of iron (III ) sulfate

now

100g of Fe gives=400/112*100=357g of iron (III ) sulfate

\(\:\)

B is your answer.!

Option c) is  correct . The cuvette will not be same as last week or spectrophotometer may give different results on different days.

What is calibration plot?

This curve is used in analytical chemistry for the determination the concentration of an unknown sample solution.

Some errors might get during measurements

1. didn't wait for instrument to calibrate. (Before showing 0.00 have placed cuvette in it)

2. Not handled the cuvette in correct way.

3. May changed the cuvette. (Size of cuvette)

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Chlorine is more reactive than bromine because it replaces both bromine and iodine.

Fluorine is the most sensitive while on the other hand, the astatine is the least reactive as compared to other elements. The chlorine displaces both bromine and iodine, and bromine displaces iodine because of its high reactivity. The element that replaces other atom is considered as more reactive.

The order of reactivity is that the chlorine is more reactive than bromine, which indicates that chlorine is more reactive than iodine.

So we can conclude that chlorine is more reactive than bromine due to high reactivity.

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Answer:

\(Q=9.2x10^5J\)

\(t=614s=10.2min\)

Explanation:

Hello,

In this case, we can compute the energy by using the following formula for air:

\(Q=nCp\Delta T\)

Whereas the moles of air are computed via the ideal gas equation at room temperature inside the 5.5m x 6.5m x 3.0m-room:

\(n=\frac{PV}{RT}\\\\V=5.5m*6.5m*3.0m=107.25m^3*\frac{1000L}{1m^3}=107250L\\ \\n=\frac{1atm*107250L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\n=4386.8mol\)

Now, we are able to compute heat, by considering that the temperature raise is given in degree Celsius or Kelvins as well:

\(Q=4386.8mol*21\frac{K}{mol*K}*10K \\\\Q=9.2x10^5J\)

Finally, we compute the time required for the heating by considering the heating rate and the required heat, shown below:

\(t=\frac{9.2x10^5J}{1.5\frac{kJ}{s}*\frac{1000J}{1kJ} } \\\\t=614s=10.2min\)

Regards.

The change in temperature from start to finish for the air and each of the samples would be calculated by subtracting the initial temperature from the final temperature. For example, if the initial temperature of the air was 20°C and the final temperature was 15°C, the change in temperature would be -5°C.

The water sample that lost the least amount of heat energy over the 6-minute time interval would be the one with the smallest change in temperature. In the example above, if the change in temperature for water sample 1 was -3°C and the change in temperature for water sample 2 was -4°C, water sample 1 would have lost the least amount of heat energy.

The rate of heat loss was not constant during each experiment because the slope of the lines on the graph is not constant. The slope of the line represents the rate of heat loss, and if the slope is not constant, the rate of heat loss is not constant.

The material that was the best insulator would be the one with the smallest change in temperature. In the example above, water sample 1 would be the best insulator because it had the smallest change in temperature.

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Answer: near gale

Explanation:

Answer:

melting

Explanation:

Answer:

Explanation:

When looking at the periodic table, the period (row) corresponds to the number of shells or PEL (principled energy level).

Hydrogen is in the first period and thus has one shell.

\(K_{sp}\) of the compound is found to be  5.04 ×\(10^{-10}\).

Solubility :Solubility can be define as the amount of a substance that dissolves or mixes in a given amount of solvent at specific conditions.

Solubility equilibrium

Ksp = \([A^{+} ]^{a}\) \([B^{-} ]^{b}\)

Ksp = solubility product constant

A+ = cation in an aquious solution

B- = anion in an aqueous solution

a, b = relative concentrations of a and b

Given,

Solubility = s = 3.60 × \(10^{-2}\) g/L

molar mass = 288 g/ mol

∴ s= 3.60 × \(10^{-2}\) g/L ÷ 288 g/ mol = 1.25 ×\(10^{-4}\) mol/ L

Reaction:

MX3 ⇄ M + 3X

           s       3s

\(K_{sp}\) =[ \(M^{+3}\)] [ \(X^{-1}\)\(]^{3}\) = solubility product

∴ \(K_{sp}\) =\([s]^{} [3s]^{3}\)

∴ \(K_{sp}\) = 3 \(s^{4}\)

∴ \(K_{sp}\) = 3 × (3.60 × \(10^{-2}\) \()^{4}\)

∴ \(K_{sp}\) = 503.8848 ×\(10^{-8}\)  = 5.04 ×\(10^{-10}\)

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The equilibrium constant for the system at this temperature is\(1.17 × 10^-31 mol^2/L^2\).

For the chemical equation:

N2(g) + O2(g) ⇌ 2NO(g)

The equilibrium mixture at a temperature of 1500 K is determined to contain 6.4 × 10^-3 mol/L of N2,\(1.7 × 10^-3\)mol/L of O2 and 1.1 × 10^-5 mol/L of NO.  First, we need to calculate the concentration of N2 and O2 required to produce

1.1 × 10^-5 mol/L of NO:

2NO(g) = N2(g) + O2(g)

Given that there are 1.1 × 10^-5 mol/L of NO, the number of moles of N2 and O2 are equal since the stoichiometric ratio is 1:1. Therefore:

\(1.1 × 10^-5 mol/L\) = [N2][O2]Kc = \(([NO]^2)/([N2][O2])Kc\)= \((1.1 × 10^-5 mol/L)^2/(6.4 × 10^-3 mol/L)(1.7 × 10^-3 mol/L)Kc\) =

1.17 × 10^-31 mol^2/L^2.

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Answer:

985.2 moles of nitrogen can produce 1970.4 moles of ammonia.

Explanation:

The balanced chemical equation for the production of ammonia from nitrogen is:

N2 + 3H2 → 2NH3

From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

So, to determine how many moles of ammonia can be produced from 985.2 moles of nitrogen, we need to use the mole ratio from the balanced chemical equation as follows:

985.2 moles N2 x (2 moles NH3 / 1 mole N2) = 1970.4 moles NH3

Therefore, 985.2 moles of nitrogen can produce 1970.4 moles of ammonia.

0.72 moles SO2 contains 4.33 × \(10^{23}\) molecules.

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When copper is placed in water, it reacts with the water molecules to form copper(II) ions and hydrogen gas. The reaction is exothermic, which means it releases heat energy into the surroundings. By measuring the temperature changes that occur, we can determine the amount of heat that is released by the reaction.

The temperature changes can be measured using a thermometer. We can place the copper metal in a container of water and take the initial temperature reading. Then, we can add the copper to the water and record the temperature change over time. By monitoring the temperature changes, we can observe the exothermic reaction taking place.

The heat released by the reaction between copper and water has many practical applications, including in the design of power plants and in the production of steam for heating and electricity generation. Therefore, understanding the heat released during this reaction is important for a variety of scientific and engineering fields.

In conclusion, step 7 of putting copper metal in water and measuring the temperature changes allows us to observe and measure the heat released by the exothermic reaction between copper and water, which has important applications in various scientific and engineering fields.

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Answer:

Aluminum

copper

Iron

Lead

The Final Slide:

Aluminum- 0.90

Copper- 0.35

Iron- 0.44

Lead- 0.12

Explanation:

I hope this helps! :))))

Answer:

Polyamides

(not entirely sure if this is right)

The pH at the equivalence point in the titration of 50.0 ml of 0.300 M CH₃COOH with 0.3 M NaOH is 8.61.

At the equivalence point in the titration of CH₃COOH with NaOH, the pH is determined by the hydrolysis of the salt formed, which is CH3COO⁻ and Na+.

The equation for the hydrolysis reaction is:

CH3COO⁻ + H2O ⇌ CH3COOH + OH⁻
We can use the Ka value for CH₃COOH to find the Kb value for CH3COO⁻ using the equation:

Kw = Ka x Kb

Where Kw is the ion product constant for water (1.0 x 10⁻¹⁴).
Kb = Kw/Ka

= (1.0 x 10⁻¹⁴)/(1.8 x 10⁻⁵)

= 5.56 x 10⁻¹⁰
Now we can use the Kb value to find the concentration of OH⁻ at the equivalence point using the equation:

Kb = ([OH⁻][CH3COOH])/[CH3COO⁻].

Since the concentrations of CH3COOH and CH3COO⁻ are equal at the equivalence point, we can simplify the equation to Kb = ([OH⁻]²)/[CH₃COO⁻].
Hence,
[OH-] = \(\sqrt{Kb(CH3COO-)}\)

=\(\sqrt{5.56(10^{-10})(0.3) }\)

= 4.06 x 10⁻⁶ M
Use the concentration of OH⁻ to find the pH at the equivalence point using the equation pOH = -log[OH-] and the relationship pH + pOH = 14.
pOH = -log(4.06 x 10⁻⁶)

= 5.39
pH = 14 - 5.39

= 8.61
Therefore, the pH at the equivalence point in the titration of 50.0 ml of 0.300 M CH3COOH with 0.3 M NaOH is 8.61.

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Answer:

Genetic Diversity Examples

Genetic diversity is the total number of genetic characteristics in the genetic makeup of a species, it ranges widely from the number of species to differences within species and can be attributed to the span of survival for a species.

Explanation:

i hope this helps u.