Convert 25.4 grams of barium phosphate, Ba3(PO4)2 to formula units.

Use Boyle's Law: P₁V₁ = P₂V₂. Since pressure and volume are inversely related and we're increasing the pressure, we should expect the new volume to be less than 2.60 liters.

Here, P₁ = 1.00 atm, V₁ = 2.60 liters, and P₂ = 3.50 atm; we want to find the new volume, V₂, at P₂. We can modify the equation to solve for V₂:

V₂ = P₁V₁/P₂ = (1.00 atm)(2.60 L)/(3.50 L) = 0.714 L

So the volume of the gas at a pressure of 3.50 atm will be 0.743 L.

Answer:

Cost.  

Weapons Proliferation Risk.  

Meltdown Risk.  

Mining Lung Cancer Risk.

Explanation:

Hope this helped!!!

Answer:

I believe it is C?

Sorry fi it doesn't help!

Answer:

A five-sided shape is called a pentagon.

Explanation:

Answered by NONE other than the ONE & ONLY #QUEEN herself aka #DRIPPQUEENMO!!!

Hope this helped!!!

Answer:

a pentagon

Explanation:

I remember that a pentagon has 5 sides

Answer:

Explanation: It is already known that 1 mole of the gas( or 32g of O2) is equivalent to 22.4 Litres of the oxygen gas. So, 8g is equivalent to = (22.4/32) × 8 = 5.6 L of the gas.

Answer:

I'd say 624.2^3 inches.

Explanation:

Explanation:

Carbon has the ability to form long chains of carbon atoms as well as ringed compounds by continuously bonding to itself. This covalent bonds may be single, double, or triple. Carbon forms covalent bonds easily with other elements, especially hydrogen, oxygen, nitrogen, halogens, and a variety of nonmetals.

Answer:   First we need to know how many moles of each reactant there are.

C6H12O6 :         25g/180.06g/mol=0.1388mol

O2:                    40g/32g/mol=1.25mol

The equation tells us we need 6 moles of O2 for every 1 mole of Glucose.

6 x 0.1388 = 0.8328

So, we have more O2 then needed – it is in excess.

Glucose is the limiting reagent – we use this for the calculation.

The equation tells us we make 6 moles of CO2 for every 1 mole of Glucose

So, we make 0,8328 moles of Carbon Dioxide

Explanation:

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

Where pH is the pH of the buffer,

pKa is -log Ka = 3.25

And [NaNO₂] [HNO₂] could be taken as the moles of each compound.

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

Where [OH⁻] = [HNO₂] = x

[NaNO₂] = 0.075M

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

Answer:

Evaporation happens when a liquid substance becomes a gas. When water is heated, it evaporates. The molecules move and vibrate so quickly that they escape into the atmosphere as molecules of water vapor. ... Once water evaporates, it also helps form clouds

Answer:

605.6mg of quinine

Explanation:

Based on Lambert-Beer's law, the intensity of an optical measurement is directely proportional to its concentration.

Unknown concentration gives a reading of 327

100 ppm gives 180 of intensity

The concentration of the diluted quinine tablet is:

327 * (100ppm / 180) = 181.67 ppm. The final dilution

The concentration of the first diluted solution is:

181.67 ppm * (100.0mL / 15mL) = 1211ppm

ppm could be defined as mass of solute (In this case of quinine) in mg per liter of solution. That is:

1211mg / L

As the tablet was diluted to 500mL = 0.5L, the mass of the quinine is:

0.5L * (1211mg / L) =

Answer:

10.5 g

Explanation:

Step 1: Given data

Step 2: Calculate the moles of solute (n)

Molarity is equal to the moles of solute divided by the liters of solution.

M = n/V

n = M × V

n = 0.243 mol/L × 0.580 L = 0.141 mol

Step 3: Calculate the mass corresponding to 0.141 moles of KCl

The molar mass of KCl is 74.55 g/mol.

0.141 mol × 74.55 g/mol = 10.5 g

1.267 x 10 ^ 24 is the total number of atoms

Answer:

40 litres

Explanation:

using Boyle's law V1P1=V2P2

V1=20 l

P1= 60atm

P2= 30 atm V2=?

substituting we will have that

20×60=V2×30

V2={20×60}/30

V2=40 l

Answer:

[tex]30^{\circ}\text{C}[/tex]

Explanation:

To know the temperature at which KCl dissolves in water we need to refer to the general solubility curves.

In the case of [tex]KCl[/tex], [tex]35\ \text{g}[/tex] of it will dissolve in [tex]100\ \text{g}[/tex] of water at a minimum temperature of [tex]30^{\circ}\text{C}[/tex].

So, the the minimum temperature needed to dissolve 35 grams of KCl in 100 grams of water is [tex]30^{\circ}\text{C}[/tex].

Answer:

jhdgafhgafhagfhafg

Explanation:

Answer:

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Water moves through the hydrosphere in a cycle. Water collects in clouds, then falls to Earth in the form of rain or snow. This water collects in rivers, lakes and oceans. Then it evaporates into the atmosphere to start the cycle all over again.

Answer:

A decrease in temperature decreases the number of collisions between molecules . The reaction rate will  decrease

Explanation:

Answer:

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These questions all involve special cases of the ideal gas law, namely Boyle's, Charles', and Gay-Lussac's Laws. The ideal gas law relates together the absolute pressure (P), volume (V), the absolute temperature (T), and number of moles (n) of a gas by the following:

PV = nRT

where R is the universal gas constant.

The special cases of the ideal gas law are obtained by holding constant all but two of the variables of a gas.

Boyle's Law relates the pressure and volume of a given mass of gas at a constant temperature: PV = k or P₁V₁ = P₂V₂.

Charles' Law relates the volume and temperature of a given mass of gas at a constant pressure: V/T = k or V₁/T₁ = V₂/T₂.

Gay-Lussac's Law relates the pressure and temperature of a given mass of gas at a constant volume: P/T = k or P₁/T₁ = P₂/T₂.

Depending on what we're given and instructed to find in each question, we can figure out which law to use.

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Question 2:

We are given the volume of a gas at some pressure, and we're to find the new volume of the gas at a different pressure. Here, we use Boyle's Law: P₁V₁ = P₂V₂ where P₁ = 60 atm, V₁ = 20.0 L, and P₂ = 30 atm. We want to find V₂, which we can determine by rearranging the equation into the form V₂ = P₁V₁/P₂. Note that pressure and volume are inversely related according to Boyle's Law; since we're decreasing the pressure, the new volume of the gas should be greater than the initial volume of 20.0 L.

V₂ = (60 atm)(20.0 L)/(30.0 atm) = 40.0 L.

So, at 30 atm, the balloon will have a volume of 40.0 L.

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Question 3:

This is another Boyle's Law question. The standard pressure (our initial pressure) is 1 atm. Here, we are decreasing the volume of the gas, and we want to find the new pressure; the pressure of the gas should thus increase proportionally (the pressure will be greater than 1 atm). Rearranging Boyle's Law to solve for P₂, we get P₂ = P₁V₁/V₂.

P₂ = (1 atm)(8.00 L)/(3 L) = 2.67 atm.

So, the new pressure of the gas is 2.67 atm (or 3 atm if we're considering V₂ to comprise one significant figure).

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Question 4:

Here, we are increasing the temperature of a gas at a known pressure, and we want to determine what the new pressure will be. This is a Gay-Lussac's Law question; from the law, we see that pressure and temperature are directly proportional. Since we're increasing the temperature of the gas, we should expect the pressure of the gas to be greater than the initial 200 atm. Gay-Lussac's Law rearranged to solve for P₂ gives us P₂ = P₁T₂/T₁. When working with gas laws, temperatures must be in Kelvin (°C + 273.15 = K). So, T₁ = 300.15 K, T₂ = 350.15 K, and P₁ = 200 atm.

P₂ = (200 atm)(350.15 K)/(300.15 K) = 233 atm.

So, if the temperature is increased from 27 to 77 °C, the pressure of the gas in the tennis ball will be 233 atm. Here, it's ambiguous how many sig figs to use; if we use one sig fig per P₁, then our P₂ would equal P₁, which I think would be an absurd for a question to ask for. I would stick with either 233 atm or 230 atm (following the two sig figs of the temperatures), or you may go with however you've been instructed.

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Question 5:

This is a Charles' Law question; we're looking for the new volume of a gas when the temperature of the gas is increased. As was the case in Gay-Lussac's Law, the two parameters in Charles' Law—volume and temperature—are directly proportional. Since the temperature of the gas is increased, we should expect the new volume of the gas to also increase (V₂ will be greater than 5.00 L). Temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (5.00 L)(300.15 K)/(250.15 K) = 5.99 L.

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Question 6:

Another Charles' Law question. As with question 5, we want to find the new volume of the gas after a change in temperature. This time, the final temperature is lower than the initial temperature, so we should expect that V₂ will be less than the initial 0.5 L. Again, temperatures in Kelvin.

V₂ = V₁T₂/T₁ = (0.5 L)(313.15 K)/(493.15 K) = 0.317 L.

So, the volume of the balloon when it is fully cooled by your refrigerator will be 0.317 L.

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Question 7:

This is yet another Charles' Law question, and, again, we are solving for V₂ after a change in temperature. Since the final temperature is greater than the initial temperature, V₂ should be greater than 2.2 L. Again, the temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (2.2 L)(653.15 K)/(453.15 K) = 3.17 L.

The new volume of the gas is 3.17 L ≈ 3.2 L (two sig figs).

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Question 8:

We return to Gay-Lussac's Law here; pressure and temperature are directly proportional, and the temperature of the gas is increased. Thus, P₂ should be greater than 3 atm. Again, remember that temperatures must be in Kelvin.

P₂ = P₁T₂/T₁ = (3 atm)(298.15 K)/(288.15 K) = 3.1 atm.

So, the pressure inside the can after the temperature rise is 3.1 atm. Not a big increase, but an increase nonetheless.

Answer:the correct answer is a.

Answer:

There are 1.8021⋅1024 molecules of CH4 in 48 grams of CH4 .

Explanation:

hope it helps you

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Answer:

a. Aqueous

Explanation:

Took the test

Explanation:

[tex]6.15 \: mol \: oxygen \: \times \frac{2 \: mol \: c2h10}{9 \: mol \: oxygen} = 1.37 \: moles \: of \: c2h10 \: needed [/tex]

If it was helpfull, please select as brainliest answer. thanks;)

Answer:

deez cutz

Explanation:

did i get it right

Answer:

Iwill add 2 for Na and 2 for NaCl so the answer is "C"