Determine the hydroxide ion concentration in
a solution that is 0.00034 M Ca(OH)2.
Answer in units of M.

Answer:

C. some substances can move freely across the cell membrane, while others must be transported.

Explanation:

Answer: Tadpole, also called polliwog, aquatic larval stage of frogs and toads. Compared with the larvae of salamanders, tadpoles have short, oval bodies, with broad tails, small mouths, and no external gills. The internal gills are concealed by a covering known as an operculum.

Explanation:

Answer:

55.0g water can be made

Explanation:

To solve this question, we must convert the molecules of H2 to moles using Avogadro's constant. With the moles, and the reaction:

H2 + 1/2O2 → H2O

We can find the moles of H2O = Moles H2 and its mass of using molar mass of water -H2O = 18.01g/mol-

Moles H2 = Moles H2O:

1.84x10²⁴ molecules * (1mol / 6.022x10²³ molecules) = 3.055 moles H2O

Mass:

3.055 moles H2O * (18.01g / mol) = 55.0g water can be made

Answer:

I dont know

Explanation:

good luck

Answer:

13%

Explanation:

Answer:

194g of distilled water.

Explanation:

Hello there!

In this case, according to the given information for this problem, it turns out possible for us to use the given mass of the solution and the percent by mass of NaF to firstly calculate the grams of this solute as shown below:

[tex]\%m=\frac{m_{solute}}{m_{solution}} *100\%\\\\m_{solute}=\frac{\%m*m_{solution}}{100\%} \\\\m_{solute}=\frac{3.0\%*200.0g}{100\%} \\\\m_{solute}=6g[/tex]

And finally, since the mass of solution is calculated by adding mass of solute and mass of solvent we obtain the mass of water (solvent) as follows:

[tex]m_w=200g-6g=194g[/tex]

Therefore, the answer is 194g of distilled water

Regards!

Answer:chemical formula of acetic acid is  or

so, molecular mass of acetic acid = 2 × atomic mass of C + 4 × atomic mass of H + 2 × atomic mass of O

= 2 × 12 + 4 × 1 + 2 × 16

= 24 + 4 + 32

= 60g/mol

given mass of acetic acid = 22g

so, no of moles of acetic acid = given mass/molecular mass

= 22/60 ≈ 0.367

so, number of moles of acetic acid is 0.367mol

number of molecules in 0.367 mol of acetic acid = 6.022 × 10²³ × 0.367

= 2.21 × 10²³

Explanation:

Answer:

316.227766

Explanation:

Answer:

[tex]V_2=8.4L[/tex]

Explanation:

Hello there!

In this case, according to the definition of the Boyle's law, which describes de pressure-volume behavior as an inversely proportional relationship, it is possible for us to write:

[tex]P_1V_1=P_2V_2[/tex]

Thus, since we are given the initial pressure and temperature, and the final pressure, we are able to calculate the final volume as shown below:

[tex]baV_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{8.0L*1.013bar}{ 0.968bar}\\\\V_2=8.4L[/tex]

Regards!

Answer: D.) non-spontaneous.

Explanation:

Answer:

258.71 dm³

Explanation:

We'll begin by calculating the number of mole of O₂ produced by the decomposition of 12 moles of KClO₃. This can be obtained as follow:

The balanced equation for the reaction is given below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 12 moles of KClO₃ will decompose to produce = (12 × 3)/2 = 18 moles of O₂.

Finally, we shall determine the volume of the O₂. This can be obtained as follow:

Temperature (T) = 325 K

Pressure (P) = 188 KPa

Number of mole (n) = 18 moles

Gas constant (R) = 8.314 KPa.dm³/Kmol

Volume (V) =?

PV = nRT

188 × V = 18 × 8.314 × 325

188 × V = 48636.9

Divide both side by 188

V = 48636.9 / 188

V = 258.71 dm³

Thus, 258.71 dm³ of oxygen were obtained from the reaction.

Answer: The pressure is 1137.5 mm Hg its pressure if its volume is increased to 150 [tex]cm^{3}[/tex] at 35 °C

Explanation:

Given: [tex]P_{1}[/tex] = 750 mm Hg,    [tex]V_{1} = 130 cm^{3}[/tex],     [tex]T_{1} = 20^{o}C[/tex]

[tex]P_{2}[/tex] = ?,          [tex]V_{2} = 150 cm^{3}[/tex],            [tex]T_{2} = 35^{o}C[/tex]

Formula used to calculate the new pressure is as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{750 mm Hg \times 130 cm^{3}}{20^{o}C} = \frac{P_{2} \times 150 cm^{3}}{35^{o}C}\\P_{2} = 1137.5 mm Hg[/tex]

Thus, we can conclude that the pressure is 1137.5 mm Hg its pressure if its volume is increased to 150 [tex]cm^{3}[/tex] at 35 °C.

Answer:

H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .

Explanation:

When

H2O2 acts as an oxidizing agent

H2O2 + 2e- 2H+--->   2H2O

Reducing agent

H2O2 --> O2 + 2e + 2H+

H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .

Answer:

1.5 × 10³ g

Explanation:

Step 1: Given and required data

Step 2: Calculate the temperature change

ΔT = 28.5°C - 22.0 °C = 6.5 °C

Step 3: Calculate the mass (m) of water

We will use the following expression.

Q = c × m × ΔT

m = Q / c × ΔT

m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g

Answer:

https://www.clutchprep.com/chemistry/practice-problems/70217/hi-aq-h2o-l-h3o-aq-i-aq-identify-each-as-either-a-bronsted-lowry-acid-bronsted-l

Explanation:

https://www.clutchprep.com/chemistry/practice-problems/70217/hi-aq-h2o-l-h3o-aq-i-aq-identify-each-as-either-a-bronsted-lowry-acid-bronsted-l

The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].

To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.

The balanced chemical equation for the reaction between solid potassium and chlorine gas is:

2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)

From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.

First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:

moles of K = 15.20 g / 39.10 g/mol = 0.388 mol

moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol

Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:

Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl

Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl

We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.

To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.

We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:

moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])

                                = 0.080 mol K

This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:

mass of K needed = 0.080 mol K x 39.10 g/mol

                              = 3.13 g K

Therefore, The 3.13 g of K would be needed to completely react with the remaining.

Example verification:

Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?

Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]

Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol

Moles of additional [tex]Cl_2[/tex] = 0.0070 mol

The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:

0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl

Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:

5.95 g + 1.04 g = 6.99 g

The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].

Learn more about Solid Potassium at

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Answer:

hope you liked it!!!!!!

A compound is a pure substance that is composed of elements chemically bonded in definite proportions. A compound can be broken down into simpler substances only by chemical reactions, such as electrolysis.

A solution is a homogeneous mixture, meaning that it is the same throughout. A solution is composed of one or more solutes dissolved in a solvent. The proportions of the solute(s) can vary, as the components of a solution are not chemically bonded. The components of a mixture can be separated by physical means, such as filtration and distillation

Answer:

[tex]\frac{250}{3}[/tex]

Explanation:

you can expect to get a 3 (theoretically) 1 time every 6 times you roll. A 1/6 chance.

Here's the equation:

[tex]\frac{1}{6} =\frac{x}{500}[/tex]

cross multiply (i think that's what it is called)

500=6x

divide by 6 on both sides:

x=[tex]\frac{250}{3}[/tex] or approx 83 times.

Hope this helps! Lmk if u have more questions <3

Answer:

(J) All organisms are composed of cells

Answer:

A

Explanation:

brainliest pls

Answer: C3H8 + 5O2 = 3Co2 +4H2O

Explanation: Equations must be balanced

You must have the same amount of C

H and O on both sides of the equation

Answer:

It will go faster each time because she is stirring therefore the water can get to the salt faster than it just sitting at the top

Explanation:

The final temperature of the gas is 269.9 K, or -3.25 °C, which was calculated with the help of ideal gas equation.

The ideal gas equation gives the relation between pressure, volume and temperature.

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

To find the initial volume of the gas. We can use the fact that the number of moles of gas does not change during the cooling process:

n = 0.850 moles

We can also use the ideal gas law to find the initial volume of the gas:

PV = nRT

V = nRT/P

where R = 0.08206 L atm/K mol is the gas constant.

Convert the initial temperature from Celsius to Kelvin:

T1 = 85 °C + 273.15 = 358.15 K

Substitute the given values into the equation:

V₁ = (0.850 mol)(0.08206 L atm/K mol)(358.15 K)/(1 atm) = 24.03 L

Now we can use the combined gas law to find the final temperature:

(P₁V₁/T₁) = (P₂V₂/T₂)

Substitute the given values into the equation:

(1.25 atm)(24.03 L)/(358.15 K) = (P₂) (17.55 L)/(T2)

Solve for T₂:

T₂ = (P₂)(17.55 L)(358.15 K)/(1.25 atm)(24.03 L)

T₂ = 269.9 K

Therefore, the final temperature of the gas is 269.9 K, or -3.25 °C.

Learn more about ideal gas equation, here:

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Answer:

One movement that i can't do is float in mid air

Explanation:

Answer:

matter is anything that occupies space

states of matter : solid,liquid, gas,plasma

Answer:

matter can be anything, tables chairs, literally anything. it has volume and takes up space.

Explanation:

Solids, liquids, gases, plasmas, and Bose-Einstein condensates (BEC)

Answer:

The main function of the cell membrane is to protect the cell from the outer environment.

Answer: The answer is D.) to separate the inside of the cell from the outside environment

Explanation:

Answer:

[tex]C_{alloy}=0.497\frac{J}{g\°C}[/tex]

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

[tex]Q_{allow}=-(Q_{water}+Q_{Al})[/tex]

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

[tex]m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})[/tex]

Then, we solve for specific heat of the metallic alloy to obtain:

[tex]C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}[/tex]

Thereby, we plug in the given data to obtain:

[tex]C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}[/tex]

Regards!