How many moles of potassium iodide, KI, are required to precipitate all of the lead (II) ion from 25.0 mL of a 1.6 M Pb(NO3)2 solution? (First, write a balanced equation for the reaction.)

Answer:

310K

Explanation:

Rearrange PV=nRT to get T=PV/nR

T=(2.0L)(0.75atm)/(0.5mol)(0.08206)

=36.5 or 37

add 273 for K to get

310

Answer:

Model C has more double the oxygen than A. I think it is B

Explanation:

Answer:

Brass is an alloy, and either a "solid solution".

Alloys in general may be solid solutions or they simply be mixtures

Explanation:

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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Answer:

i dont speak mexican

Explanation:

In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.

When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

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Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]

Thus, the density of the calcite sample will be:

[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

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Regards!

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

Let's consider the following unbalanced redox reaction.

I⁻ + SO₄²⁻ → I₂ + H₂S

The corresponding half-reactions are:

I⁻ → I₂

SO₄²⁻ → H₂S

We will perform the mass balance adding OH⁻ and H₂O where appropriate.

2 I⁻ → I₂

6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Then, we will perform the charge balance adding electrons where appropriate.

2 I⁻ → I₂ + 2 e⁻

8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Finally, we will multiply the first half-reaction by 4 and the second by 1, and add them.

4 × (2 I⁻ → I₂ + 2 e⁻)

1 × (8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻)

------------------------------------------------------------

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

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Answer: the answer is 0.074Hz

Explanation:

Given, (In air)

Velocity V=340m/s

Frequency f=20,000Hz

Wavelegth λ=?

V=f.λ

λ=

F

V

​

=

20,000

340

​

=0.017Hz

Also, Given (in Water)

Velocity, V=1480m/s

Frequencyf=20,000Hz

wavelength, λ=?

V=F.λ

λ=

F

V

​

=

20,000

1480

=​0.074Hz

Answer:

frequency

Explanation:

frequency is velocity/ wavelength

340/1.36

250

Answer:

The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.

Explanation:

Answer:

dinitrogen pentachloride

Answer:

Explanation:

The material that is not a mixture; it has the same properties all the way through is called a substance. Thus the material that is not a mixture; it has the same properties all the way through is called a substance.

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The partial pressure of oxygen is equal to 10.2 atm.

Given the following data:

To determine the partial pressure of oxygen:

First of all, we would find the total number of moles of the elements:

[tex]n=3.25+2.75[/tex]

n = 6 moles

Next, we would determine the mole fraction of the oxygen by using this formula:

[tex]Molefraction \;of \;a \;substance =\frac{No.\; of \; moles \;of \;substance}{Total \;no. \;of \; moles \;of \;substances}[/tex]

Substituting the values, we have:

[tex]Molefraction \;of \;a \;substance =\frac{2.75}{6} \\\\Molefraction \;of \;a \;substance =0.4583[/tex]

For oxygen:

[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.4583 \times 22.4[/tex]

Partial pressure of oxygen = 10.2 atm.

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Answer:

12.1

Explanation:

3.25 + 2.75= 6 moles total

then take: 3.25/6 to get 0.542

multiply 0.542 by 22.4 to get 12.1

the other answer solved for nitrogen, not oxygen.

Explanation:

since 5g saturates 12.5g of water at 10°c

so......x would saturate 50g of water at 0°c

then you can cross multiply

Answer:

Tetrahedral molecules are normally spy hybridized.

Explanation:

The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Cr

But:

1 mole of Cr = 52 g

Thus, we can say that:

6.02×10²³ atoms = 52 g of Cr

With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:

6.02×10²³ atoms = 52 g of Cr

Therefore,

5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³

5.15×10²¹ atoms = 0.44 g of Cr

Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

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Answer:

11000 g/L

Explanation:

Since density is mass divided by volume, find the volume of the metal ball by subtracting the initial volume of water in the graduated cylinder from the ending volume. In this case, the volume would be 17.90 mL - 16.05 mL =  1.85 mL = 0.00185 L. Then using the formula for density, calculate 20.350 g / 0.00185 L = 11000 g/L.

Density is an intensive property as it does not depend on the quantity of the substances Whereas mass and volume are extensive property. Therefore, the density of the metal ball is 11000 g/L.

Density tells about the compactness of the substances, how much dense is the substances in other words. Object that is more denser than water they just sink in the water.

Mathematically,

Density = Mass of the metal ball ÷volume marked on the graduated cylinder

volume=ending volume- initial volume of water in the graduated cylinder            

            =17.90 mL - 16.05 mL

            =  1.85 mL

            = 0.00185 L.

substituting all the given values in the above equation, we get

density=  20.350 g / 0.00185 L

            = 11000 g/L.

Therefore, the density of the metal ball is 11000 g/L.

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Answer:

i can't understand the question

One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.

Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.

We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.

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Answer:

78n

Explanation:

The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.

The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.

The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.

Given, the input force = 15 N

The mechanical advantage of the hammer = 5.2

Mechanical advantage = Output force/ Input force

5.2 = Output/15

Output force = 15 ×5.2 = 78 N

Therefore, the force is exerted on the nail in the claw is equal to 78 N.

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Explanation:

a+b =ab

answer

ab added all

is give answer

Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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Answer:

16.7%

Explanation:

40 ml of salt water + 200 ml of solution = 240 ml

40/240 = 4/24 = 1/6=16.7%