What is the relative humidity of the air when the dry-bulb temperature is 4°C and the dewpoint is -4°C?
1.
42%
2.
46%
3
51%
4
56%

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:

Fe is limiting reactant and 3.00g of Fe2O3 will be produced

Explanation:

To solve this question we must convert the mass of each reactant to moles and, using the reaction we can find limiting reactant. With moles of limiting reactant we can find moles of Fe2O3 and its mass as follows:

Moles Fe -Molar mass: 55.845g/mol-

2.1g * (1mol / 55.845g) = 0.0376 moles

Moles O2 -Molar mass: 32g/mol-

2.1g * (1mol / 32g) = 0.0656 moles

For a complete reaction of 0.0656 moles of O2 are needed:

0.0656moles O2 * (4mol Fe / 3 mol O2) = 0.0875 moles Fe

As there are just 0.0376 moles,

Fe is limiting reactant

The mass of Fe2O3 is:

Moles:

0.0376 moles Fe* (2mol Fe2O3 / 4mol Fe) = 0.0188 moles Fe2O3

Mass:

0.0188 moles Fe2O3 * (159.69g / mol) =

3.00g of Fe2O3 will be produced

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:

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

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:

–272.96 °C

Explanation:

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

Initial temperature (T₁) = 27.0 °C

Initial volume (V₁) = 630 L.

Final volume (V₂) = 92.0 mL

Final temperature (T₂) =?

Next, we shall convert 27.0 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 27.0 °C

Initial temperature (T₁) = 27.0 °C + 273

Initial temperature (T₁) = 300 K

Next, we shall convert 92.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

92 mL = 92 mL × 1 L / 1000 mL

92 mL = 0.092 L

Next, we shall determine the final temperature.

Initial temperature (T₁) = 300 K

Initial volume (V₁) = 630 L.

Final volume (V₂) = 0.092 L

Final temperature (T₂) =?

V₁ / T₁ = V₂ / T₂

630 / 300 = 0.092 / T₂

2.1 = 0.092 / T₂

Cross multiply

2.1 × T₂ = 0.092

Divide both side by 2.1

T₂ = 0.092 / 2.1

T₂ = 0.04 K

Finally, we shall convert 0.04 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

Final temperature (T₂) = 0.04 K

Final temperature (T₂) = 0.04 – 273

Final temperature (T₂) = –272.96 °C

Answer:

382.49 C degree Celsius

Explanation:

Hello,

This problem deals with understanding the ideal gas law which hopes to predict how ideal gases might behave in any given condition. I listed the formula below and we are basically just going to solve for temperature by rearranging the equation as seen on the picture (there's also other rearranged ones in case you need to solve for those).

Universal gas constant R has a value of 0.0821 L * atm/(mole * K) when working with these given units so it will be part of this equation. R value changes based on what units you have.

T = PV/nR

   = (2.33) (72.31) / (3.13)(0.0821)

   = 655.64 K

Question is asking temperature in celsius so we employ the formula attached below:

C = K - 273.15

   = 655.64-273.15

    = 382.49 degree Celsius

382.49 degree Celsius is the answer!

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:

Answer:

316.227766

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!

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:

I dont know

Explanation:

good luck

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:

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

Explanation:

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: D.) non-spontaneous.

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:

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:c it’s dropped off in the kidneys

Explanation:

I took the quiz

The solid wastes are from the circulatory system is expelled to the kidney where, the nitrogenous wastes like urea and uric acid is excreted as urine from the body.

A circulatory system is an organ system, where the blood is purified and oxygenation of blood takes place. Through circulatory system, the blood reaches throughout the body pumped from the heart through veins.

The organs included in circulatory system are lungs, heart, aorta, veins, blood vessels etc. There are various kinds of blood vessels each having specific functions.

There is a network of blood vessels including arteries and large veins, capillaries that join the venules and other veins. All the nutrients and ions are circulated throughout the body through blood and solid wastes are then expelled to kidney.

Kidney function as a sieve to clean the good fluid from waste products. Uric acid and urea along with water excreated as urine then. Thus, option C is correct.

To find more about circulatory system, refer the link below:

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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:

(J) All organisms are composed of cells

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: 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:

(a) 157.7 g

(b) 7.04 g/dm³

Explanation:

(a) From the question,

According to Avogadro's Law,

1 mole of every gas at STP occupies a volume of 22.4 dm³

But mass of 1 mole of the diatomic gas  = molar mass of the gas.

This Implies that,

The molar mass of the gas at STP occupies a volume of 22.4 dm³

From the question,

If,

0.25 L bottle contain 1.76 g of the gas,

Therefore,

Molar mass of the gas = (1.76×22.4)/0.25

Molar mass of the gas = 157.7 g.

(b) Density of the gas = mass/volume

D = m/v

Given: m = 1.76 g, v = 0.25 L = 0.25 dm³

Therefore,

D = 1.76/0,25

D = 7.04 g/dm³

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

Answer:

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

Explanation: