Examining the equations or equilibrium constants related to a base, salt, or an acid is an indirect way to determine strength of an electrolyte. The strength of an electrolyte can be examined directly by placing a solution into a circuit so that the voltage or amount of current can be measured. Although conductivity refers to the flow of charged species, we usually examine conductivity with respect to resistance. As the name implies, a solution that does not conduct electricity very well also has a very high resistance. If solutions containing various acids, bases, and salts were prepared and connected to a circuit that powers a light bulb, the strength of the electrolyte could be estimated by examining the intensity of the light bulb. Complete the following sentences regarding the experimental determination of the electrolyte strength for various molecules dissolved in solution.
a. strong electrolyte
b. weak electrolyte
c. non-electrolyte
1. A 20 mL solution containing 2 mmol of Ca3(PO4)2(s) was integrated into to a circuit that powers a light but, When the power supply was turned on, the light bulb produced a glow. Ca3(PO4)2(s) is__.
2. A 20 mL, solution containing 2 mmol of C12H22O11(aq) was integrated into a circuit that powers a light bulb. When the power supply was turned on the light bulb remained off. C12H22O11(aq) is___.
3. A 20 mL, solution containing 2 mmol of HF(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb faintly flickered. HF(g) is an____.
4. A 20 mL solution containing 2 mmol of LiOH(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb produced a bright glow LiOH(s) is a____.

Answer:

[tex]m_{H_2}=12.6gH_2[/tex]

Explanation:

Hello there!

In this case, given the reaction:

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

It is possible to compute the necessary grams of hydrogen which produce 71.1 g of ammonia, given the 3:2 mole ratio and their molar mass as shown below:

[tex]m_{H_2}=71.1gNH_3*\frac{1molNH_3}{17.04gNH_3}*\frac{3molH_2}{2molNH_3} *\frac{2.02gH_2}{1molH_2}\\\\m_{H_2}=12.6gH_2[/tex]

Best regards!

Answer:

B) Transportation and photosynthesis

C) Cellular respiration and photosynthesis

Answer: Its D the guy above is wrong dont use his answer he got me a 75

Explanation:

Answer:

Number three, because precipitation is rain, which comes from cumulonimbus clouds. Precipitation happens when clouds have accumulated enough evaporated water for the cloud to get heavy and need to get rid of some of the water, which it does through condensing the water and then delivering it as rain.

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Answer:

No

No

Yes 25cm

Explanation:

1) mass can't be divided by distance.

2) the measurements are completely different but under the same category of distance. you can only divide it if meters was cm and squared

3) measurements the same and you can multiply it

Answer:

0.415 j/g°C

Explanation:

Given that :

Mass of water, m1= 250.0 g

Temperature of water, T1= 25°C

Specific heat capacity of water, C= 4.184 J/g°C

Mass of brass , m2= 51.1g

Temperature of brass = 95.4°C

Specific heat capacity of brass , c2=?

Final temperature = 26.4°C

Heat lost by brass = Heat gained by water

mc(dT) = mc(dT)

51.1 * c * (95.4 - 26.4) = 250 * 4.184 * (26.4 - 25)

51.1 * 69 * c = 250 * 4.184 * 1.4

3525.9c = 1464.4

c = 1464.4 / 3525.9

c = 0.4153265

c = 0.415 j/g°C

Specific heat capacity of Brass for the experiment = 0.415 j/g°C

Answer: [tex]MgS[/tex], [tex]MgBr_2[/tex], [tex]V_2S_5[/tex] and [tex]VBr_5[/tex] are four binary ionic compounds that could be formed.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Mg is having an oxidation state of +2 called as [tex]Mg^{2+}[/tex] cation and [tex]S^{2-}[/tex] is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]MgS[/tex]. Also [tex]Mg^{2+}[/tex] cation and [tex]Br^{-}[/tex] can combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]MgBr_2[/tex].

Also Vanadium is having an oxidation state of +5 called as [tex]V^{5+}[/tex] cation and [tex]S^{2-}[/tex] is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]V_2S_5[/tex]. Also [tex]V^{5+}[/tex] cation and [tex]Br^{-}[/tex] anion can combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]VBr_5[/tex].

Answer:

0.43 atm

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 0.95 atm

Initial volume (V₁) = 0.55 L

Final volume (V₂) = 1.22 L

Final pressure (P₂) =?

The final pressure of the gas can be obtained by using the Boyle's law equation as follow:

P₁V₁ = P₂V₂

0.95 × 0.55 = P₂ × 1.22

0.5225 = P₂ × 1.22

Divide both side by 1.22

P₂ = 0.5225 / 1.22

P₂ = 0.43 atm

Therefore, the final pressure of the gas is 0.43 atm

The final pressure of the gas is 0.43 atm.

Pressure is the process of force exerted perpendicular to any surface.

Given, the Initial pressure (P₁) is 0.95 atm

The initial volume (V₁) is 0.55 L

The final volume (V₂) is 1.22 L

To find the final pressure (P₂)

By the Boyle's Law

[tex]\rm P_1V_2= P_2V_2[/tex]

[tex]\rm 0.95\times 0.55 = P_2\times1.22\\\\0.5225 = P_2 \times 1.22\\\\Divided\;both \;side\;by\;1.22\\\\P_2=\dfrac{0.5225}{1.22} = 0.43 atm.[/tex]

Thus, the final pressure of the gas is 0.43 atm.

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Answer: The periodic table is a chart where different types of elements organized by what they have in common.

Explanation:

Answer:

Sexual Reproduction

Explanation:

In Asexual Reproduction offsprings are know as ramets and many ramets are known as clowns. So, when two parents are involved genetic variations occur in offspring. Pollen possess male gamete (antherozoid/sperm) and ovule possess egg.

A part

Answer:

sdasdasddsafsadsfds

Explanation:

Answer:

B.) P4

Explanation:

Step 1: Convert 25.0 grams of phosphorus into moles of phosphorous:

25 g P4 x 1 mol P4/123.90 g P4 = 0.202 mol P4

Step 2: Convert 0.202 mol P4 into moles of O2 using their stoichiometric ratios:

0.202 mol P4 x 5 mol O2/1 mol P4 = 1.01 mol O2

Step 3: Convert 1.01 mol O2 into grams of O2:

1.01 mol O2 x 31.98 g O2/1 mol O2 = 32.3 g O2

Because 25.0 grams of phosphorous only needs 32.3 grams of oxygen to react, phosphorous will be the limiting reactant, as after the 25.0 grams of phosphorous is used up, there will still be 17.7 grams of oxygen leftover (excess reactant).

Because the reaction cannot occur without more phosphorous to react with the 17.7 grams of oxygen remaining, P4 is the limiting reactant.

Answer:  B.) P4

Explanation:

        25 g P4 x 1 mol P4/123.90 g P4 = 0.202 mol P4

        0.202 mol P4 x 5 mol O2/1 mol P4 = 1.01 mol O2

        1.01 mol O2 x 31.98 g O2/1 mol O2 = 32.3 g O2

Because 25.0 grams of phosphorous only needs 32.3 grams of oxygen to react, phosphorous will be the limiting reactant, as after the 25.0 grams of phosphorous is used up, there will still be 17.7 grams of oxygen leftover (excess reactant).

Because the reaction cannot occur without more phosphorous to react with the 17.7 grams of oxygen remaining, P4 is the limiting reactant.

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Answer:

B

Explanation:

you can tell by looking at the image lol the wires have to be connected

(brainliest please? c:)

( although its fine if not im just glad to have helped )

Answer:

The answer is B

Explanation:

Answer:

carbon

Explanation:

I hope this helps :)

Answer:

carbon

Explanation:

YOU CAN MARK ME AS BRAINIEST IF YOU WANT

Answer:

13.9kj/mol

Explanation:

[tex]h = 891 -(393.5 +2*241.8)=1768.1=13.9kj/mol[/tex]

Answer:

[tex]0.21\ \text{kg/s}[/tex]

Explanation:

P = Pressure = [tex]0.48\ \text{atm}=0.48\times 101325\ \text{Pa}[/tex]

V = Volume = [tex]450\ \text{L/s}=450\times 10^{-3}\ \text{m}^3/\text{s}[/tex]

R = Gas constant = [tex]8.314\ \text{J/mol K}[/tex]

T = Temperature = [tex](264+273.15)\ \text{K}[/tex]

The reaction is

[tex]2H_2S+3O_2\rightarrow 2SO_2+2H_2O[/tex]

From ideal gas equation we have

[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{0.48\times101325\times 450\times 10^{-3}}{8.314\times (264+273.15)}\\\Rightarrow n=4.9\ \text{mol}[/tex]

Moles of [tex]SO_2[/tex] produced is

[tex]\dfrac{2}{3}\times 4.9=3.267\ \text{moles}[/tex]

Molar mass of [tex]SO_2[/tex] = 64.066 g/mol

Production rate is

[tex]3.267\times 64.066=209.3\ \text{g/s}=0.21\ \text{kg/s}[/tex]

The rate at which sulfur dioxide is being produced [tex]0.21\ \text{kg/s}[/tex].

Answer:

First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volumes in liters, 'N' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmosphere per moles Kelvin).

Explanation:

I majored in Chemistry

The question is missing some parts. Here is the complete question.

Carbon monoxide is a colorless, odorless gas that binds irreversibly to hemoglobin in our blood, causing suffocation and death. CO is formed during incomplete combustion of carbon. One way to represent this equilibrium is:

[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]

we could also write this reaction three other ways listed below. The equilibrium constant for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex]       K₁ =

2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex]    K₂ =

3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex]    K₃ =

Answer: 1) [tex]K_{1}=\frac{1}{K}[/tex]

              2) [tex]K_{2}=\frac{1}{K^{1/2}}[/tex]

             3) [tex]K_{3}=K^{1/2}[/tex]

Explanation: A chemical reaction can be reversible, i.e., can proceed in both directions: to the right of the arrow (forward) or towards the left of the arrow (backward).

When the rates of forward and backward reactions are the same, the reaction is in equilibrium. In that state, we can determine the equilibrium constant, [tex]K_{c}[/tex].

For the first way to represent equilibrium of CO formed, the [tex]K_{c}[/tex] is calculated

[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]

[tex]K=\frac{[O_{2}]}{[CO]^{2}}[/tex]

in which the symbol [ ] is concentration of the compound.

In equilibrium constant, solids are not included.

Equilibrium constants for the other reactions:

1) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex]  

[tex]K_{1}=\frac{[CO]^{2}}{[O_{2}]}[/tex]

Comparing K₁ and K, the first one is the inverse of K, so writing in terms of K

[tex]K_{1}=\frac{1}{K}[/tex]

2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex]

[tex]K_{2}=\frac{[CO]}{[O_{2}]^{1/2}}[/tex]

In terms of K, K₂ is

[tex]K_{2}=\frac{1}{K^{1/2}}[/tex]

3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex]

[tex]K_{3}=\frac{[O_{2}]^{1/2}}{[CO]}[/tex]

This constant in terms of K will be

[tex]K_{3}=K^{1/2}[/tex]

In conclusion, K₁, K₂ and K₃ in terms of K is [tex]\frac{1}{K}[/tex],[tex]\frac{1}{K^{1/2}}[/tex] and [tex]K^{1/2}[/tex], respectively.

Answer:

0.093 M

Explanation:

First, we assume that the density of household hydrogen peroxide is 1 g/mL. We also assume that there's no reaction between hydrogen peroxide and sodium carbonate, if there were, in the end all hydrogen peroxide would be consumed.

Now we calculate how many grams of H₂O₂ were added, using the given volume, concentration and density:

We convert grams to moles:

0.45 g H₂O₂ ÷ 34 g/mol = 0.013 mol H₂O₂

Now we divide the number of moles by the final volume, to calculate the molarity of H₂O₂:

Based on the given concentration of the original hydrogen peroxide solution, the molarity of the hydrogen peroxide (in mol/L) of the final solution is 0.093 M.

Assuming that the density of household hydrogen peroxide is 1 g/mL and that no reaction occurs between hydrogen peroxide and sodium carbonate.

The mass in grams of H₂O₂ added is first determined using the given volume, concentration and density:

mass = 15 mL *  1 g/mL * 3g/100g  = 0.45 g H₂O₂

Moles of H₂O₂ is determined as follows:

molar mass of H₂O₂ = 34 g/mol

moles of H₂O₂ = 0.45 g H₂O₂ / 34 g/mol

moles of H₂O₂ = 0.013 mol H₂O₂

molarity of H₂O₂ is then calculated as follows:

final volume = 15 mL + 120 mL + 5 mL = 140 mL

final volume = 140 mL / 1000 = 0.140 L

molarity = 0.013 mol H₂O₂ / 0.140 L

molarity of H₂O₂ = 0.093 M

Therefore, the molarity of the hydrogen peroxide (in mol/L) of the final solution is 0.093 M.

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Answer:

Pretty sure it's C

Explanation:

Answer:  [tex]X_3Z_4[/tex]

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here metal X is having an oxidation state of +4 called as [tex]X^{4+}[/tex] cation and [tex]Z^{3-}[/tex] is an anion with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]X_3Z_4[/tex]

[tex]X_3Z_4[/tex] is the formula of the resulting compound.

ANSWER:

Nonpolar

sorry if I'm wrong

Answer:

I think radiant I’m not sure

Explanation: