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Answer:

Measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water

Explanation:

To make 100 mL of 0.001 mM solution from 0.0405mM solution, we need to determine the volume of 0.0405mM solution needed. This can be obtained as follow:

Molarity of stock (M₁) = 0.0405 mM

Volume of diluted (V₂) = 100 mL

Molarity of diluted solution (M₂) = 0.001 mM

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

0.0405 × V₁ = 0.001 × 100

0.0405 × V₁ = 0.1

Divide both side by 0.0405

V₁ = 0.1 / 0.0405

V₁ = 2.47 mL

Therefore, to make 100 mL of 0.001 mM solution from 0.0405mM solution, measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water.

Answer:

d- The resulting pH will be equal to 7 because a strong base will neutralize a strong acid.

Explanation:

The reaction between potassium hydroxide and hydrochloric acid of equal volume and equal concentration yields a solution of pH 7 at equivalence point. We must note that KOH is a strong base while HCl is a strong acid. This fact influences the pH of the system at equivalence point.

Owing to the fact that the acid is exactly neutralized by the base; at the equivalence point of such titration, it is expected that hydrogen ions(H+) and hydroxide ions (OH-) must have reacted to form water, this leads to a final pH of 7.  

Answer:

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Explanation:

thanks for the points ❤️

Answer:

[tex]P_2=16atm[/tex]

Explanation:

Hello there!

In this case, according to the given information, we can infer we need to use the Gay-Lussac's equation it order to understand this pressure-temperature relationship as shown below:

[tex]\frac{P_1}{V_1}=\frac{P_2}{V_2}[/tex]

Thus, we solve for P2 and plug in P1, T1 and T2, to obtain:

[tex]P_2=\frac{P_1T_2}{T_1}\\\\P_2=\frac{8atm*600K}{300K}\\\\P_2=16atm[/tex]

Regards!

A.

Particles and energy in a system become more and more disordered.

Answer:

34.44 kg

Explanation:

First we convert 1.00 kg of phosphorus (P) into moles, using its molar mass:

Then we convert 0.03125 kmoles of P into kmoles of Ca₃(PO₄)₂:

Now we calculate the mass of 0.0625 kmoles of Ca₃(PO₄)₂:

Finally we calculate the required mass of the ore, using the definition of content percentage:

Answer:

22.67 L of PH₃

Explanation:

The balanced equation is:

[tex]P_4 (s) + 6H_2(g) \to 4PH_3(g)[/tex]

From the equation:

[tex]34 L \times \dfrac{1 \ mol \ of H_2 }{22.4 \ L \ H_2} \times \dfrac{4 \ mol \ of \ PH_3}{6 \ mol \ H_2} \times \dfrac{22.4 \ L \ PH_3}{1 \ mol \ PH_3}[/tex]

= 22.67 L of PH₃

Answer:

[tex]T_F=-4.4\°C[/tex]

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to calculate the freezing point of an aqueous solution of potassium sulfide by using the following equation:

[tex]T_F=T_{solv}-i*m*Kf[/tex]

In such a way, we firstly calculate the molality of this solution according to:

[tex]m=\frac{\frac{13g}{110.262 g/mol} }{0.150kg} =0.79m[/tex]

Finally, since the Van't Hoff's factor for K2S is 3, the freezing point of the solution turns out to be:

[tex]T_F=0\°C-3*0.79m*1.86\°C/m\\\\T_F=-4.4\°C[/tex]

Regards!

Answer:

True

Explanation:

Answer:

evaporation is the process

Answer:

Explanation:

Evaporation is the process of becoming a vapor. The process of extracting moisture as by heat.

Evaporation is the process of change from a liquid to a gas at temperatures below the boiling point.

Therefore, the final answer is evaporation.

Explanation:

[tex]H _{2}O _{2(aq)} →H _{2}O _{(l)} + O _{2}(g) \\ solution : 2 \: and\: 2[/tex]

Answer:

six cookies?(1 cal _4.184 j; ALSO: 1 Calorie -1 kilocalorie)

Answer:

True

Explanation:

The tertiary structure of a protein is a complex arrangement formed as the polypeptide chain folds and twists.

This folding & twisting of polypeptide chain leading to its complex structure, is true about tertiary structure of protein. It occurs due to different interactions between side chains of amino acids.

Answer:

Combustion

Explanation:

Answer:

A

Explanation:

edge21

Explanation:

Answer:

Sis I think it happened with me but I am not able to remember if u want u can share if it happened with u

Answer:

hot water is thinner than cold water.

Explanation:

The rate of diffusion of a substance has a lot to do with the temperature of a body.

Let us take water for instance, cold water has a greater density than hot water. As a result of this, the molecules in cold water are slower when in motion and more sluggish.

On the other hand, the molecules of hot water are quite faster since hot water has a lower density(thinner than cold water).

Thus, an  increase in temperature will cause an increase in diffusion rate because hot water is thinner than cold water.

Answer:

B, No reaction will occur

Explanation:

Copper as compared to lead is less reactive. This is the reason when  lead is added to copper nitrate solution, it replaces the copper and itself combines with nitrate to form lead nitrate aqueous solution

Lead + Copper(II) nitrate → Copper + Lead (II) nitrate

The same is not the case when the reaction is revered i.e Cu is added to Pb NO3 solution.

Hence, option B is correct

Answer:

Any element in group 18 has eight valence electrons (except for helium, which has a total of just two electrons

Solution :

Given :

Hydrogen iodide decomposes to hydrogen and iodine gas

    [tex]$2 HI \ \ \ \Leftrightarrow \ \ \ \ H_2 \ \ \\ + \ \ I_2 $[/tex]

I    0.148            0          0

C     -2a           +a          +a

E  0.148-2a      a            a

We know

[tex]$k_p=\frac{P(H_2)P(I_2)}{P(HI)^2}$[/tex]

[tex]$0.016=\frac{a^2}{(0.148-2a)^2}$[/tex]

[tex]$0.016^{1/2}=\frac{a}{0.148-2a}$[/tex]

[tex]$0.12649=\frac{a}{0.148-2a}$[/tex]

0.0187 = 1.2529 a

a = 0.0149

Therefore

P(HI) = 0.148 - 2a

        = 0.148 - 2(0.0149)

        = 0.1182 atm

P([tex]$H_2$[/tex]) =  a

        = 0.0149 atm

P([tex]$I_2$[/tex]) =  a

        = 0.0149 atm

Answer:d

Explanation:

Answer:

When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood.

When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood.

Carbonic acid can raise or lower the pH of blood.

Explanation:

A buffer is a solution that resists changes to its pH when small quantities of acids or bases are added to it. The human blood serves as a buffer as it contains a buffer of carbonic acid (H2CO3) and bicarbonate anion (HCO3-) which serves to maintain blood pH between 7.35 and 7.45. Other buffering systems in blood exist such as the Hydrogen ion and oxygen gas which affects oxygen binding to haemoglobin, however the carbonic-acid-bicarbonate buffer is the most important buffer for maintaining acid-base balance in the blood.

A buffer solution is made up of an acid and its conjugate base or a base and its conjugate acid. For carbonic acid-bicarbonate buffer, carbonic acid serves as the acid while bicarbonate serves as the base. When a little quantity of a base as hydroxide ions is added to a buffer, the acid reacts with it and remove it from the solution. On the other hand, when a little quantity of an acid as hydrogen ions are added to a buffer, the conjugate base reacts with it and remove it from the solution, thus keeping the pH of the solution fairly constant.

In the carbonic acid-bicarbonate buffer:

When blood is too basic, carbonic acid can ionize to bicarbonate and H+ ions, adding H+ ions to the blood.

When blood becomes too acidic, bicarbonate combines with extra H+ ions to form carbonic acid, removing H+ ions from the blood.

Thus, carbonic acid can raise or lower the pH of blood.

Carbonic acid work to maintain blood pH as follows:

This means that, carbonic acid works to maintain blood pH as follows:

Learn more about buffers at: https://brainly.com/question/24188850

Answer: [tex]K_{eq}[/tex] at  the temperature of the experiment is 0.56.

Explanation:

Moles of  [tex]CO[/tex] = 0.35 mole

Moles of  [tex]H_2O[/tex] = 0.40 mole

Volume of solution = 1.00 L

Initial concentration of [tex]CO[/tex] = [tex]\frac{0.35mol}{1.00L}=0.35M[/tex]

Initial concentration of [tex]H_2O[/tex] = [tex]\frac{0.40mol}{1.00L}=0.40M[/tex]

Equilibrium concentration of [tex]CO[/tex] = [tex]\frac{0.19mol}{1.00L}=0.19M[/tex]

The given balanced equilibrium reaction is,

                      [tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)[/tex]

Initial conc.          0.35 M         0.40 M                  0 M        0M

    At eqm. conc.     (0.35-x) M   (0.40-x) M   (x) M      (x) M

Given:  (0.35-x) = 0.19

x= 0.16 M

The expression for equilibrium constant for this reaction will be,

[tex]K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}[/tex]  

Now put all the given values in this expression, we get :

[tex]K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}[/tex]

[tex]K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56[/tex]

Thus [tex]K_{eq}[/tex] at  the temperature of the experiment is 0.56.

Answer:

A. Yes

B. The new temperature of the gas is -116 °C

Note: The question is incomplete. The complete question is given below :

For many purposes we can treat butane C H10) as an ideal gas at temperatures above its boiling point of - 1. °C. Suppose the pressure on a 500 mL sample of butane gas at 41.0°C is cut in half. Iyes Is it possible to change the temperature of the butane at the same time such that the volume of the gas doesn't change? yes no If you answered yes, calculate the new temperature of the gas. Round your answer to the nearest °C.

Explanation:

A. According to the pressure law of gases,for a fixed mass of gas the pressure of a gas is directly proportional to its Kelvin temperature once the volume is kept constant. This means that a change in temperature can bring about a change in pressurein a gas at constant volume.

B. From the pressure law of gasese: P1/T1 = P2/T2

Where initial pressure = P1, final pressure = P2

Initial temperature = T1, final temperature = T2

For the butane gas;

P1 = P

P2 = P/2

T1 = 41°C = (273 + 41 ) K = 314 K

T2 = ?

From the equation, T2 = T1 × P2 / P1

T2 = 314 × P/2 /P

T2 = 157 K

T2 = (157 - 273) °C = -116 °C

Therefore, the new temperature of the gas is -116 °C

I'm gonna guess E on this one, but I think you should choose either E or A

Answer:

D. 331 mm Hg

Explanation:

We can solve this problem by keeping in mind the law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of its components.

In other words:

We input the given data:

And calculate the pressure of the nitrogen:

Answer:

chlorine ~ 30%

zinc ~ 28%

oxygen ~ 41%

Explanation: