Someone please help me

Answer:

Calculate the pH of 0.010 M HNO2 solution. The K, for HNO2 is 4.6 x 104

Answer: pH = 2.72

Answer:

8 moles

Explanation:

When we are asked to convert from grams of a substance into moles, we have to use the substance's molar mass.

Meaning that for this problem, we'll use the molar mass of hydrogen peroxide (H₂O₂), as follows:

There are 8 moles in 272 grams of hydrogen peroxide.

Answer:

Explanation:

pretty sure its gaillieo

Answer:

Galileo

Explanation:

he was the first one to use a telescope

Answer:

C

Explanation:

thermal energy is the another name for heat

Answer:

b

Explanation:

i know

Answer:

Explanation:

From the information given:

Mass of BiPO₄ = 0.3451 g

Number of moles of BiPO₄ = [tex]0.3451 \ g \ BiPO_4 \times \dfrac{1 \ mol \ BiPO_4}{303.95 \ g \ BiPO_4}[/tex]

[tex]= 0.001135 \ mol[/tex]

The number of moles of Bi³⁺ in 20.00 mL is:[tex]= 0.001135 \ mol \ BiPO_4 \times \dfrac{1 \ mol \ of \ Bi^{3+}}{1 \ mol \ BiPO_4}[/tex]

=  0.001135 mol of Bi³⁺

The number of moles of Bi³⁺ in 100 mL stock solution

[tex]= 0.001135 \ mol \ Bi^{3+} \times \dfrac{100 \ mL}{20.0 \ mL}[/tex]

[tex]= 0.005675 \ mol[/tex]

Mass of BSS in 4.9993 g tablets

[tex]m = 0.005675 \ mol \ Bi^{3+} \times \dfrac{1 \ mol \ BSS}{1 \ mol \Bi^{3+}} \times \dfrac{362.1 \ g \ BSS}{1 \ mol \ BSS}[/tex]

m = 2.055 g BSS

Mass of BSS in 5.0103 g (5 tables)

[tex]m = 2.055 g \ BSS \times \dfrac{5.0103 \ g}{4.9993 \ g}[/tex]

= 2.06 g

∴

The mass of BSS per tablet is [tex]=\dfrac{2.06 \ g}{5 \ tablet}[/tex]

= 0.412 g BSS/ tablet

Answer:

A. PbI2(s) ===> Pb2+(aq) + 2I-(aq)

B. Ksp = [Pb2+][I-]^2

C. 1.52 x 10^-3 M. It is equal to the moles/L of PbI2 that go into solution.

D. 2 x 1.52 x 10^-3 = 3.04 x 10^-3 M

E. Ksp = (1.52x10^-3)(2.31x10^-6) = 3.51 x 10^-9

Explanation:

In the given question, [tex]\rm PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)[/tex] is the dissolution reaction of [tex]\rm PbI_2[/tex] in water,   [tex]\rm Ksp = [Pb^{2+}][I^-]^2[/tex] is expression for Ksp for [tex]\rm PbI_2[/tex],  [tex]1.52\times 10^{-3 }[/tex] mol/L is the concentration of [tex]\rm Pb^{2+ }[/tex] in the equilibrium solution,  [tex]3.04\times 10^{-3}[/tex] mol/L is the concentration of I- in the equilibrium solution and

[tex]1.40\times 10^{-8}[/tex] is the solubility product of  [tex]\rm PbI_2[/tex], respectively.

A reaction is a process that involves the transformation of one or more substances into one or more different substances.

1. The dissolution reaction of [tex]\rm PbI_2[/tex] in water is:

[tex]\rm PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)[/tex]

2. The expression for Ksp for [tex]\rm PbI_2[/tex] is:

 [tex]\rm Ksp = [Pb^{2+}][I^-]^2[/tex]

Where, [tex]\rm [Pb^{2+}][/tex] is the concentration of [tex]\rm Pb^{2+ }[/tex] ions in solution and [tex]\rm [I^-][/tex] is the concentration of [tex]\rm I^-[/tex] ions in solution.

3. The solubility of  [tex]\rm PbI_2[/tex] is [tex]1.52\times 10^{-3 }[/tex] mol/L. Since [tex]\rm PbI_2[/tex] dissociates into one  [tex]\rm Pb^{2+ }[/tex] ion and two [tex]\rm I^-[/tex] ions, the concentration of  [tex]\rm Pb^{2+ }[/tex] in solution is equal to the solubility of  [tex]\rm PbI_2[/tex], which is [tex]1.52\times 10^{-3 }[/tex] mol/L.

[tex]\rm [Pb^{2+}] = 1.52\times 10^{-3}\ mol/L[/tex]

4. Since  [tex]\rm PbI_2[/tex] dissociates into one [tex]\rm Pb^{2+ }[/tex] ion and two  [tex]\rm I^-[/tex] ions, the concentration of  [tex]\rm I^-[/tex] ions in solution is twice the solubility of  [tex]\rm PbI_2[/tex] .

[tex]\rm [I^-] = 2 \times 1.52\times 10^{-3 }\ mol/L[/tex]

= [tex]3.04\times 10^{-3}[/tex] mol/L

5. The solubility product of  [tex]\rm PbI_2[/tex] can be calculated using the expression for Ksp and the concentrations of [tex]\rm Pb^{2+ }[/tex] and  [tex]\rm I^-[/tex] ions in solution.

[tex]\rm Ksp = [Pb^{2+}][I^-]^2[/tex]

=  [tex]1.52\times 10^{-3 }[/tex]  [tex]\times[/tex] [tex]\rm (3.04\times 10^{-3} mol/L)^2[/tex]

= [tex]1.40\times 10^{-8}[/tex]

Therefore, the dissolution reaction of [tex]\rm PbI_2[/tex] in water, expression for Ksp for [tex]\rm PbI_2[/tex], the concentration of Pb2+ in the equilibrium solution, the concentration of I- in the equilibrium solution and solubility product of  [tex]\rm PbI_2[/tex] is mentioned above.

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Answer:

Learn how gifted students think. ...

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Explore real-word application.

Explanation:

Sonic is my bf forever, back off

Answer:

oof I always lose on video games so I just quit

Answer:

Explanation:

The answer is true.

Answer:

true

Explanation:

acids have ph which is less than 7 and base have ph greater than 7

Answer:

c substance

Explanation:

Matter is never destroyed or created in chemical reactions. The particles of one substance are rearranged to form a new substance. The same number of particles that exist before the reaction exist after the reaction.

Answer:

calcium has more mass

Explanation:

1 mole calcium mass= 40.078 g

1 mole magnesium mass= 24.305 g

Answer:

4. KCl(s)

Explanation:

KCl is an ionic salt that dissolves in water to form a KCl aqueous solution.

A solution is defined as the homogeneous mixture of one or more solutes dissolved in a solvent. Here in the saturated solution of potassium chloride, the solute is KCl. The correct option is 4.

A solute is defined as the substance which is dissolved in a solution. In a solution the amount of the solute present is always smaller than the amount of the solvent. For example in a salt solution, salt dissolves in water and therefore salt is the solute.

The particles of the solute present in a solution cannot be seen by our eye. The solute from a solution is not possible to separate by filtration. In an unsaturated solution, the concentration of the solute is much lower than that of the concentration of the solvent.

A solution is a combination of the solute and the solvent.

Thus the correct option is 4 - KCl.

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Answer:

a.

[tex]rate_{NH_3}=\frac{2d[NH_3]}{dt}[/tex]

b.

[tex]rate_{N_2}=\frac{-1d[N_2]}{dt} \\\\rate_{H_2}=\frac{-3d[H_2]}{dt}[/tex]

Explanation:

Hello there!

In this case, according to the law of rate proportions, it is possible to write the the rates of reaction for the formation of NH3 and the disappearance of the N2 and H2, by considering that the coefficient in the reaction for NH3 is +2 and those of N2 and H2, -1 and -3 respectively. Moreover, we set up these equations as derivatives as shown below:

a.

[tex]rate_{NH_3}=\frac{2d[NH_3]}{dt}[/tex]

b.

[tex]rate_{N_2}=\frac{-1d[N_2]}{dt} \\\\rate_{H_2}=\frac{-3d[H_2]}{dt}[/tex]

Best regards!

Answer:

The agricultural revolution increased agricultural production and technological advancements. I think this led to population growth because the increase in labor and the increase in technologies increased human deveolpment. The revolution also allowed farmers to grow and produce more food and transport it to where it was needed.

Explanation:

Answer:

c. 10–14[H3O+]

Explanation:

On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34). Example: What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10-5 M? The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH.

Answer:

0.875

...............

Answer: 1. 2.7 moles  of ammonia are formed

2. 12.0 moles  of hydrogen are required

3. 2.0 moles of nitrogen are required

Explanation:

The balanced chemical equation is:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

According to stoichiometry:

3 moles of hydrogen form = 2 moles of ammonia

Thus 4.0 moles of hydrogen form =[tex]\frac{2}{3}\times 4.0=2.7moles[/tex] of ammonia

According to stoichiometry:

2 moles of ammonia are formed by = 3 moles of hydrogen

Thus 8.0 moles of ammonia are formed by  =[tex]\frac{3}{2}\times 8.0=12.0moles[/tex] of hydrogen

According to stoichiometry:

3 moles of hydrogen react with = 1 mole of nitrogen

Thus 6.0 moles of hydrogen react with =[tex]\frac{1}{3}\times 6.0=2.0moles[/tex] of nitrogen

Answer:

15.2 g H2

Explanation:

2H2O -> 2H2 + O2

9.06 x 10^24 molecules x (1 mol/6.022 x 10^23 molecules) x (2 mol H2/2 mol H2O) x (1.008 g/1 mol) = 15.2 g H2

Answer:

I believe the answer is "b". "During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis."

Explanation:

I remember from last year but I'm not totally sure. Good luck!

The experimental setup has been the design that includes the experimental and the control group. during the experiment, only one variable is changed to test the hypothesis. Thus, option b is correct.

The experiment design includes the testing of the hypothesis to solve the problem and reach the solution through the analysis of the observations. It has been the setup that includes the test and the control group that involves the dependent, independent, and controlled variables.

Testing one variable in the experimental designs means changing one of the elements of the experience to prove the reliability of the hypothesis. This variable is the factor that alters and affects the other factors of the experiment and leads to a cause and effect.

The result of the effect of the hypothesis proves the testability of the experiment and provides solutions to the research question. This further leads to the formulation of the law and theory.

Therefore, option A. testing one variable involves proving the hypothesis.

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Answer:

yayyy! proud of u

Explanation:

Answer:

Distance cover by Paul's sister by swim = 38.125 miles

Explanation:

Given:

Distance cover by Paul by swim = [tex]7\frac{5}{8}[/tex] = 61/8 miles

Distance cover by Paul's sister by swim = 5 times Distance cover by Paul

Find:

Distance cover by Paul's sister by swim

Computation:

Distance cover by Paul's sister by swim = 5 times Distance cover by Paul

Distance cover by Paul's sister by swim = 5 x [61 / 8 miles]

Distance cover by Paul's sister by swim = 305 / 8 miles

Distance cover by Paul's sister by swim = 38.125 miles

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

Producers are the foundation of every food web in every ecosystem—they occupy what is called the first tropic level of the food web. The second trophic level consists of primary consumers—the herbivores, or animals that eat plants. At the top level are secondary consumers—the carnivores and omnivores who eat the primary consumers. Ultimately, decomposers break down dead organisms, returning vital nutrients to the soil, and restarting the cycle. Another name for producers is autotrophs, which means “self-nourishers.” There are two kinds of autotrophs. The most common are photoautotrophs—producers that carry out photosynthesis. Trees, grasses, and shrubs are the most important terrestrial photoautotrophs.  In most aquatic ecosystems, including lakes and oceans, algae are the most important photoautotrophs.

Answer:

m = 176.04 g .

Explanation:

Hello there!

In this case, according to the mole-mass relationships, which are based off the mass of one mole of any compound via the molar mass, it is possible to realize that the molar mass of carbon dioxide is 44.01 (12.01+16*2) g/mol, and therefore, the mass in grams of 4.00 moles of this compound are calculated as shown below:

[tex]m=4.00molCO_2*\frac{44.01gCO_2}{1molCO_2}\\\\m=176.04g[/tex]

Best regards!

Answer:

all of the above

Explanation

Answer:

i think its the gibbous phase

Explanation:

The statement which is  given below describe a homogeneous catalyst.

"It is in the same phase as the reactants"

The catalyst which have similar phase to the reactant of the reaction is called homogeneous catalyst.

Mostly in  homogeneous catalyst, catalyst present in liquid form.

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