Describe the formation of an aqueous libr solution when solid libr dissolves in water
Fill in the blanks with words given below.
K and I atoms
K and IF ions
dissociation
atoms
KI molecules
polar dilution
hydration
molecules ions
nonpolar
At the_______ surface of the solid _____________are pulled into solution by___________ the water molecules, where the______________ process surrounds separate with water molecules.

Answers

Answer 1

Answer and Explanation:

The water is a polar solvent which dissolves into the LiBr molecules and converts into [tex]Li^+[/tex] and [tex]Br^-[/tex] ions far from the solid also into the solution, when they are hydrated.

Now, the complete words are as given below:-

The [tex]K^+[/tex] and [tex]I^-[/tex] ions at the surface of the solid are pulled into the solution by the polar water molecules, where the hydration process surrounds separate ions with water molecules.

Answer 2

LiBr has been the ionic compound and has been dissociated in the water by the force of the polar water molecules. The water molecules that have been polar in nature exert the force onto the ionic compounds and help in the dissociation.

At the dissociation surface of the solid KI molecules, are pulled into the solution by the polar water molecules, where the hydration process surrounds ions and separates them with water molecules.

For more information about the hydration process, refer to the link:

https://brainly.com/question/939764


Related Questions

Write the empirical formula

Answers

Answer:

[tex]1) NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3) NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}[/tex]

Explanation:

[tex]1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}[/tex]

Balance the following equations: (c) H2(g)+I2(s)⟶HI(s)H2(g)+I2(s)⟶HI(s)

Answers

Answer: [tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.

The balanced chemical reaction is:

[tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]

which factor is most responsible for the fact that water is a liquid rather than a gas at room temperature

Answers

The answer is Hydrogyn bonding. It keeps the water molocules bonded together and in a liquid state, without it it'd be in a gashious state.

Answer:Hydrogen bonds keep them together in room temperature, hope this helps!

Explanation:

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A pentavalent cation atom has 20 and 15 neutrons as protons. Find the electron quantity and mass number respectively. (40 pts.) a) 20 and 15 b) 15 and 20 c) 15 and 35 d) 35 and 15 e) 10 and 20

Answers

Answer:

C.

Explanation:

Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.

What volume of 6.00 M hydrochloric acid is needed to prepare 500 mL of 0.100 M solution?

Answers

Answer:

8.33mL or .0083L

Explanation:

Use m1 * V1 = m2 * V2

6.00M(x) = 0.100M(500mL)

solve for x

x= (.1 * 500) / 6

x=8.333 mL

A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is saying 91g I keep getting 175g. Can someone please explain?

Answers

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

What is the specific heat of a 85.01 g piece of an unknown metal that exhibits a 45.2°C temperature change upon absorbing 1870 J of heat?

Answers

Answer:

[tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

Explanation:

In this question, we have to remember the relationship between Q (heat) and the specific heat (Cp) the change in temperature (ΔT), and the mass (m).

[tex]Q=m*Cp*ΔT[/tex]

The next step is to identify what values we have:

[tex]Q~=~1870~J[/tex]

[tex]m~=~85.01~g[/tex]

[tex]ΔT~=~45.2~^{\circ}C[/tex]

[tex]Cp~=~X[/tex]

Now, we can plug the values and solve for "Cp":

[tex]1870~J=~85.01~g~*Cp*45.2~^{\circ}C[/tex]

[tex]Cp=\frac{1870~J}{85.01~g~*45.2~^{\circ}C}[/tex]

[tex]Cp=0.48~\frac{J}{g~^{\circ}C}[/tex]

The unknow metal it has a specific value of [tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

I hope it helps!

When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.430 g of a particular hydrocarbon was burned in air, 0.446 g of CO, 0.700 g of CO2, and 0.430 g of H2O were formed.

Required:
a. What is the empirical formula of the compound?
b. How many grams of O2 were used in the reaction?
c. How many grams would have been required for complete combustion?

Answers

Answer:

(a) The empirical formula of the compound is

m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).

(b) The grams of O2 that were used in the reaction is 1.146 g

(c) The amount of O2 that would have been required for complete combustion is 1.401 g.

Explanation:

a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)

(b) Using law of conservation of mass from above

m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)

m(O2) = 0.446 + 0.700 + 0.430 - 0.430

m(O2) = 1.146 g

The grams of O2 that were used in the reaction is 1.146 g

(c) for complete combustion, we need to oxidized CO to CO2

Then, 2CO +O2 = 2CO2

m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}

m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g

Note; Molar mass of O2 = 32, CO = 28

m(total)(O2) = m(O2) + m(add)(O2)

m(total)(O2) = 1.146 + 0.255 = 1.401 g

The amount of that grams would have been required for complete combustion is 1.401 g.

Note (add) and (total) were used subscript to "m"

a ______ consumer is a heterotroph that directly eats an autotroph. A primary B. Quaterany C tertiary D secondary

Answers

Primary consumer just remember the “primary” (first) consumer so the one that eats a heterotroph

Which element has the largest atomic radius

Answers

Answer:

Francium

Explanation:

The atomic radius increases from top to bottom in a group, and decreases from left to right across a period.

so francium (Fr) is the largest atom or has highest radii.

Hope this helps & please mark as brainiest!

Answer:

Francium has the largest atomic radius.

The general trend for atomic radii is increasing from top to bottom and decreasing from left to right so the one with the largest atomic radius will be in the bottom left of the periodic table.

At what geographical location would the boiling point of water be lowest?
A. Boston, Massachusetts
B. The coast of the Atlantic Ocean
C. The Dead Sea
D. The top of Mount Everest​

Answers

Answer:

I think it would be the Dead Sea

Explanation:

Because the dead sea is already usually in the warmer temperatures, the boiling point of the water would be lower than the rest.

C. The Dead Sea is the correct choice

Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [

Answers

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

            HCOOH ⇄ H⁺ + HCOO⁻

I               1.0M          0          0

C              -x            +x        +x

E            1.0-x            x          x

For the steps below, refer to the ICE chart above.

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

When a strip of magnesium metal is placed in an aqueous solution of copper(II) nitrate, elemental copper coats the surface of the magnesium strip and aqueous magnesium nitrate forms.
1. As a reactant, what is the charge on copper?
2. As a product, what is the charge on copper?

Answers

Answer:

1. 2+ ([tex]Cu^{2+}[/tex]).

2. 0 ([tex]Cu^0[/tex]).

Explanation:

Hello,

In this case, the described chemical reaction is a redox reaction in fact, since the oxidation states of both magnesium and copper change as shown due to the displacement:

[tex]Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)[/tex]

Therefore:

1. Since copper is the cation in the copper (II) nitrate, the (II) means that its charge is 2+ ([tex]Cu^{2+}[/tex]).

2. Since copper is alone, it means no electrons are being neither shared not given, its charge is 0 ([tex]Cu^0[/tex]).

Best regards.

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N 2O 4 and 45.0 g N 2H 4. Some possibly useful molar masses are as follows: N 2O 4 = 92.02 g/mol, N 2H 4 = 32.05 g/mol.
N 2O 4( l) + 2 N 2H 4( l) → 3 N 2( g) + 4 H 2O( g)
a) LR = N2O4, 45.7 g N2 formed
b) LR = N2O4, 105 g N2 formed
c) LR = N2H4, 13.3 g N2 formed
d) LR = N2H4, 59.0 g N2 formed
e) No LR, 45.0 g N2 formed

Answers

Answer:

Option A. LR = N2O4, 45.7g N2 formed

Explanation:

The balanced equation for the reaction is given below:

N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and mass of N2 produced from the balanced equation. This is illustrated below:

Molar mass of N2O4 = 92.02 g/mol

Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02 g

Molar mass of N2H4 = 32.05 g/mol

Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g

Molar mass of N2 = 2x14.01 = 28.02g/mol

Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g

Summary:

From the balanced equation above,

92.02g of N2O4 reacted with 64.1g of N2H4 to produce 84.06g of N2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

92.02g of N2O4 reacted with 64.1g of N2H4.

Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.

From the calculations made above, we can see that only 34.83g out 45g of N2H4 is required to react completely with 50g of N2O4.

Therefore, N2O4 is the limiting reactant and N2H4 is the excess reactant.

Finally, we shall determine the mass of N2 produced from the reaction.

In this case the limiting reactant will be used as it will produce the maximum yield of N2 since all of it is used up in the reaction.

The limiting reactant is N2O4 and the mass N2 produced can be obtained as illustrated below:

From the balanced equation above,

92.02g of N2O4 reacted to produce 84.06g of N2.

Therefore 50g of N2O4 will react to produce = (50 x 84.06)/92.02 = 45.7g of N2.

Therefore, 45.7g of N2 were produced from the reaction.

At the end of the day,

The limiting reactant is N2O4 and 45.7g of N2 were produced from the reaction.

What is the half-life for the first order decay of 14C according to the reaction, 146C — 147N +e- ?
The rate constant for the decay is 1.21 x10-4 year-1

Answers

Answer:

5727 years or 5730 (rounded to match 3 sig figs) whichever one your teacher prefers

Explanation:

First Order decay has a half life formula of Half Life = Ln (2) / k = 0.693/K

Half-life = 0.693/k = 0.693/1.21 x10-4 =  5727 years or 5730 (rounded to match 3 sig figs)

This should be correct because if you google the half-life of 14 C it is ~ 5700 years

A chemist measures the energy change Delta H during the following
2Fe2O3(s)->4FeO(s)+O2(g).
1) this reactions is: Endothermic or exothermic.
2) suppose 94.2g of Fe2O3 react. will any heat be relased or absorbed. yes absorbed. yes releases. no.
3) If you said heat will be released or absorbed in the second part of the question. calculate how much heat will be absored or released. be sure your answer has correct number of significant digits.

Answers

Answer: 1) Endothermic

2) Yes, absorbed.

3) 166.86 kJ will be absorbed.

Explanation:

1) To determine if a reaction is endothermic (heat is absorbed by the system) or exothermic (heat is released by the system), first calculate its change in Enthalpy, which is given by:

ΔH = [tex]H_{products} - H_{reagents}[/tex]

For the reaction 2Fe₂O₃(s) ⇒ 4FeO(s) + O₂(g):

Enthalpy of Reagent (Fe₂O₃(s))

Enthalpy of formation for Fe₂O₃(s) is - 822.2 kJ/mol

The reaction needs 2 mols of the molecule, so:

H = 2(-822.2)

H = - 1644.4

Enthalpy of Products (4FeO(s) + O₂(g))

Enthalpy of formation of O₂ is 0, because it is in its standard state.

Enthalpy of formation of FeO is - 272.04 kJ/mol

The reaction produces 4 mols of iron oxide, so:

H = 4(-272.04)

H = -1088.16

Change in Enthalpy:

ΔH = [tex]H_{products} - H_{reagents}[/tex]

ΔH = - 1088.16 - (-1644.4)

ΔH = + 556.2 kJ/mol

The change in enthalpy is positive, which means that the reaction is absorving heat. Then, the chemical reaction is Endothermic.

2) When Fe₂O₃(s) reacts, heat is absorbed because it is an endothermic reaction.

3) Calculate how many mols there is in 94.2 g of Fe₂O₃(s):

n = [tex]\frac{mass}{molar mass}[/tex]

n = [tex]\frac{94.2}{160}[/tex]

n = 0.6 mols

In the reaction, for 2 mols of Fe₂O₃(s), 556.2 kJ are absorbed. Then:

2 mols --------------- 556.2 kJ

0.6 mols ------------- x

x = [tex]\frac{0.6*556.2}{2}[/tex]

x = 167 kJ

It will be absorbed 167 kJ of energy, when 94.2 g of Fe₂O₃(s) reacts.

A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.

Answers

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Concentration of the weak acid (Ca): 0.187 MpH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}[/tex]

if hybridization patterns of NH and SF gave sp3 and sp3d2 respecticely what is the value of 2 x + 3 y

Answers

Answer:

The value of 2x+3y is 26

Explanation:

Because NH pattern gave hybridization Sp3

Just like NH3 Or NH2-

And SF pattern gave hybridization Sp3d2

Also just like SF6

X is the static no. of NH pattern

So X=4

Y is the static no. Of SF pattern

So Y = 6

Value of 2X+3Y

2*4 + 3*6

26

Can a catalyst change an exothermic reaction into an endothermic reaction or vice versa? Please explain your answer.

Answers

Answer:

A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.

Explanation:

Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.

Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other  releases it.

As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.

Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T

Answers

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. When 10.1 g of Mg reacts with 10.5 g of O2, 11.9 g of MgO are collected.
a) Write the balanced equation for this reaction, including all states of matter.
b) What is the limiting reactant in this reaction?
c) What is the theoretical yield for this reaction?
d) What is the percent yield for this reaction?

Answers

Answer:

A. 2Mg(s) + O2(g) —> 2MgO(s)

B. Mg is the limiting reactant.

C. Theoretical yield of MgO is 16.83g.

D. The percentage yield is 70.7%

Explanation:

A. The balanced equation for the reaction. This is given below:

2Mg(s) + O2(g) —> 2MgO(s)

B. Determination of the limiting reactant.

First, we shall determine the mass of Mg and O2 that reacted and the mass of MgO produced from the balanced equation. This is illustrated below:

Molar mass of Mg = 24g/mol

Mass of Mg from the balanced equation = 2 x 24 = 48g.

Molar mass of O2 = 16x2 = 32g/mol.

Mass of O2 from the balanced equation = 1 x 32 = 32g

Molar mass of MgO = 24 + 16 = 40g/mol

Mass of MgO from the balanced equation = 2 x 40 = 80g

Summary:

From the balanced equation above,

48g of Mg reacted with 32g of O2 to produce 80g of MgO.

Now, we can obtain the limiting reactant as follow:

From the balanced equation above,

48g of Mg reacted with 32g of O2.

Therefore, 10.1g of Mg will react with = (10.1 x 32)/48 = 6.73g of O2.

From the calculations made above, we can see that only 6.73g out of 10.5g of O2 given is needed to react completely with 10.1g of Mg.

Therefore, Mg is the limiting reactant and O2 is the excess reactant.

C. Determination of the theoretical yield of MgO.

The limiting reactant is used in this case as it will produce the maximum yield of the reaction since all of I is used up in the reaction.

The theoretical yield can be obtain as illustrated below:

From the balanced equation above,

48g of Mg reacted to produce 80g of MgO.

Therefore, 10.1g of Mg will react to produce = (10.1 x 80)/48 = 16.83g of MgO.

Therefore, the theoretical yield of MgO is 16.83g.

D. Determination of the percentage yield.

This is illustrated below:

Actual yield of MgO = 11.9g

Theoretical yield of MgO = 16.83g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 11.9/16.83 x 100

Percentage yield = 70.7%

A student wants to prepare a salt starting with H2SO4. Select all of the compound types that can react with H2SO4 to form a salt.
1. salt
2. acid
3. acid salt
4. basic oxide
5. base
6. metal
7. acidic oxide

Answers

Answer:

4 and 6 would work for this

Choose the situation below that would result in an endothermic ΔHsolution.
a) When |ΔHsolute| > |ΔHhydration|
b) When |ΔHsolute| is close to |ΔHhydration|
c) When |ΔHsolute| < |ΔHhydration|
d) When |ΔHsolvent| >> |ΔHsolute|
e) There isn't enough information to determine.

Answers

Answer:

Option A - When |ΔHsolute| > |ΔHhydration|

Explanation:

A solution is defined as a homogeneous mixture of 2 or more substances that can either be in the gas phase, liquid phase, solid phase.

The enthalpy of solution can either be positive (endothermic) or negative (exothermic).

Now, we know that enthalpy is amount of heat released or absorbed during the dissolving process at constant pressure.

Now, the first step in thus process involves breaking up of the solute. This involves breaking up all the intermolecular forces holding the solute together. This means that the solute molecules are separate from each other and the process is always endothermic because it requires energy to break interaction. Thus;

The enthalpy ΔH1 > 0.

Thus, the enthalpy of the solute has to be greater than the enthalpy of hydration.

An endothermic ΔHsolution occurs when |ΔH solute| < |ΔH hydration|.

A substance dissolves in water when the solute - solvent interaction exceeds the solute - solute solute interaction. The energy required to break the bonds between solutes is the ΔHsolute and the energy released when solute - solvent interaction take place is called the ΔHhydration.

We know that when  |ΔH solute| < |ΔH hydration|, energy is required to break up the solute - solute interaction and  ΔHsolution is endothermic.

Learn more: https://brainly.com/question/1340582

Description (with words) of water just above melting temperature. What intermolecular forces do you expect to find in water in liquid state

Answers

Answer:

intermolecular dipole-dipole hydrogen bonds

Explanation:

Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.

Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.

Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe: PbO(s) NaCl(aq) H2O(l) --> Pb(OH)Cl(s) NaOH(aq) How many grams of PbO and how many grams of NaCl would be required to produce 10.0 g of Pb(OH)Cl

Answers

Answer:

8.59 g

2.25 g

Explanation:

According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-

Moles of Pb(OH)CL is

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

[tex]= \frac{10.0 g}{259.65g / mol}[/tex]

= 0.0385 mol

Mass of PbO needed is

[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]

After solving the above equation we will get

= 8.59 g

Mass of NaCL needed is

[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]

After solving the above equation we will get

= 2.25 g

Therefore we have applied the above formula.

What is the mass of 3.45 moles
NO2?
(N = 14.01 g/mol, O = 16.00 g/mol)

Answers

Answer:

158.7 g

Its the right answer

Question 11: How does the energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital compare to the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital?

Answers

Answer:

Explanation:

The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital

Which best describes the act of using senses or tools to gather information? creating a hypothesis making an observation summarizing the results recording the measurements

Answers

Answer:

B - Making an Observation

Explanation:

Making an observation best describes the act of using senses or tools to gather information. Therefore, option B is correct.

What are senses in the scientific method?

The five senses—sight, taste, touch, hearing, and smell—gather data about our surroundings that the brain interprets. Based on prior experience (and subsequent learning), as well as by combining the data from each sensor, we make sense of this information.

Information gleaned from your five senses is referred to as an observation. These are smell, taste, touch, hearing, and sight. When you see a bird or hear it sing, you notice it.

The term observation, which is also used to sense five aspects of the world including vision, taste, touch, smell, and hearing, is used to describe utilizing the senses to examine the world, employing tools to take measurements, and looking at prior research findings.

Thus, option B is correct.

To learn more about the scientific method, follow the link:

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Each unknown mixture contains 5 metal constituents. Select the 5 metal ions that you have identified as being present in your mixture. Please double check your selections before you hit the submit button. a. Ca b. Co c. Cr d. Fe e. K f. Mn g. Zn

Answers

Explanation:

A metal ion is a type of atom compound that has an electric charge.

Such atoms willingly lose electrons in order to build positive ions called cations. The selected  Ions are :

[tex]1. Mn^2^+\\\ 2. Ca^2^+\\\ 3. Co^2^+\\\ 4. Fe^2^-\\\ 5. K^+[/tex]

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ∆G° = 349 kJ/mol

Answers

The given question is incomplete, the complete question is:

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol

A) –309 kJ/mol

B) –329 kJ/mol

C) None of the above

D) –349 kJ/mol

E) –369 kJ/mol

Answer:

The correct answer is option D, that is, -349 kJ/mol.

Explanation:

Based on the given information, the reaction is:  

NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.  

Therefore, there is a need to reverse the reaction and change the sign on ΔG.  

Now the reaction will become,  

Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).  

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