7. A student has a 25 g sample of gas in a container at STP and adds another 7

g of the gas while keeping the pressure constant, what is the new temperature in

degrees Celsius?

Answers

Answer 1

Answer:

New temperature t2 = [1.28T− 273.15]° C

Explanation:

Given:

Volume v1 = 25 gram

New volume v2 = 25 + 7 = 32 gram

Constant pressure = p

Temperature t1 = T

Find:

New temperature t2

Computation:

Pv1/t1 = Pv2/t2

25 / T = 32 / t2

t2 = 1.28T

New temperature t2 = [1.28T− 273.15]° C


Related Questions

Scientists use models to help them represent things in nature that they cannot observe directly. They can also use models to make predictions. For example, scientists use Bohr models of atoms to predict chemical reactions between atoms, You can use models in the same way, but you must first understand what each model component represents. Which element is represented by this Bohr model?

A. chlorine (Cl)
B.oxygen (O)
C. selenium (Se)
D. sulfur (S)

Answers

Answer:

I think it’s chlorine (Cl)

Explanation:

I’m literally doing the same question right now

If the earth were to spin half as fast as it does now, how many hours would be in one day?

Answers

the day would be 48 hours

Answer:

48 hours

Explanation:

Because if the world would spin regularly, then it would take 24 hours. But if it spins half as fast, then it would be 2x more.

Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts.

Answers

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³

 

Explain why the atomic radius decreases across a period and increases down a group ?​

Answers

Answer:

In general, atomic radius decreases across a period and increases down a group. ... Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.

can a water cycle ever end​

Answers

Answer:

No It cannot.

Explanation:

A water cycle does not have an end as it moves all around the earth

The process:

A water cycle can start by water evaporating through the sun's heat from a water bed, such as: Ocean, sea, pond etc.

The heat from the sun will turn the water into water vapour and will go up into the atmosphere.

The water vapor gets together with other vapors and forms clouds.

These clouds will move around the earth, once they are so full of water they drop the water to Earth in some form of precipitation. It could be rain, snow, sleet, or hail.

These water will form ponds or go back in the ocean and the process wil repeat again.

CHEMWORK The preparation of NO2(g) from N2(g) and O2(g) is an endothermic reaction: N2(g) + O2(g) + NO2(g) (unbalanced) The enthalpy change of reaction for the balanced equation (with lowest whole-number coefficients) is AH = 67.7 kJ. If 304 ml N2 (9) at 100ºC and 3.32 atm and 410 ml 0,(9) at 100°C and 3.32 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of NO2(g) ? ​

Answers

Heat required : 1.523 kJ

Further explanation

Reaction

N₂+2O₂⇒2NO₂ AH = 67.7 kJ

mol of N₂:(use Pv=nRT)

[tex]\tt n=\dfrac{3.32\times 0.304}{0.082\times 373}=0.033[/tex]

mol of O₂:

[tex]\tt n=\dfrac{3.32\times 0.41}{0.082\times 373}=0.045[/tex]

Limiting reactant :

N₂ : O₂

[tex]\tt \dfrac{0.033}{1}\div \dfrac{0.045}{2}=0.033:0.0225[/tex]

Limiting reactant : O₂(smaller ratio)

mol NO₂ = mol O₂ = 0.045

2 mol ⇒ 67.7 kJ, so for 0.045 mol, heat required :

[tex]\tt \dfrac{0.045}{2}\times 67.7=1.523~kJ[/tex]

The heat required to produce the maximum yield of NO2 is 1.52 kJ.

The balanced reaction equation is;

N2(g) + 2 O2(g) -> 2 NO2(g)

The number of moles of each gas is obtained from the ideal gas equation as follows;

For N2;

P = 3.32 atm

V = 304 ml or 0.304 L

T = 100ºC + 273 = 373 K

n = ?

R = 0.082 atmLK-1mol-1

From;

PV = nRT

n = PV/RT

n = 3.32 atm × 0.304 L/0.082 atmLK-1mol-1 × 373 K

n = 0.033 moles

For O2;

PV = nRT

n =  PV/RT

V = 410 ml or 0.410 L

P = 3.32 atm

R = 0.082 atmLK-1mol-1

T = 100°C + 273 = 373 K

n =  3.32 atm × 0.410 L/0.082 atmLK-1mol-1 × 373 K

n = 0.045 moles

1 mole of N2 reacts with 2 moles of O2

0.033 moles of N2 reacts with 0.033 moles × 2 moles/1 mole

= 0.066 moles

2 moles of O2 yields 2 moles of NO2

0.045 moles O2 yields 0.045 moles × 2 moles/2 moles = 0.045 moles

If 2 mole of NO2 requires 67.7 kJ

0.045 moles of NO2 requires 0.045 moles  × 67.7 kJ/2 mole

= 1.52 kJ

Learn more: https://brainly.com/question/9743981

At STP, 32 grams of O2 would occupy the same volume as:

Answers

At STP 32 g of O₂ would occupy by the same volume as  4 g of He

Further explanation

Complete question

At STP 32 g of O₂ would occupy by the same volume as:

4.0 g of He8.0 g of CH₄64 g of H₂32 g of SO₂

Standard Conditions

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So the gas will have the same volume if the number of moles is the same

mol of 32 grams of O₂ :

[tex]\tt \dfrac{32}{32}=1[/tex]

He

[tex]\tt mol=\dfrac{4}{4}=1[/tex]

CH₄

[tex]\tt mol=\dfrac{8}{16}=0.5[/tex]

H₂

[tex]\tt mol=\dfrac{64}{2}=32[/tex]

SO₂

[tex]\tt mol=\dfrac{32}{64}=0.5[/tex]

So mol of 4 g He = mol of 32 g O₂

STP is the standard temperature and pressure of the gas to compare with the experimental data. At STP 32 gm of oxygen is equivalent to 4 gm of helium.

What is STP?

STP is the standard temperature and pressure of 0 degrees Celsius and 1 atm. At this temperature and pressure, the volume of the gas is 22.4 L/mol.

So, if the moles of the 32 gm oxygen are the same as the other element then the volume will also be the same. Here, the moles of oxygen are 1 mol.

Moles of 4 gm helium is 1, 8 gm methane is 0.5 mol, 64 gm dihydrogen is 32 mol, and 32 gm sulfur dioxide is 0.5 mol.

Therefore, the mole of 4 gm helium is equivalent to the mole of 32 gm oxygen.

Learn more about STP here:

https://brainly.com/question/25705287

Globular proteins play a critical role in virtually all chemical processes.
True or false

Answers

Answer:

True

Explanation:

How many moles are in 1.2 x 1023 Rps?

Answers

I’m not sure what you’re talking about

hhhheeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelllllllllllllllllllllllllllllllppppppppppppppppppp​

Answers

Answer:

What do you need help with?

Explanation:

What do you need help On????

Complete and balance the equation for the single displacement reaction Cl2 + NaCl + Br2

Answers

Answer:

NaBr + Cl2 = NaCl + Br2 - Chemical Equation Balancer.

Explanation:

Jet streams generally flow toward the

a. north
b. south
c. east
d. west

please answer thank you

Answers

Answer:

The major jet streams on Earth are westerly winds that flow west to east.

i’m pretty sure the answer would be D!!!

How much kilometers is in 6 m

Answers

Answer: 0.006 km

Explanation:

divide the length value by 1000

so 6/1000 = 0.006

.006 km because 6/1000 is .006

If a cell is 80% water and the outside environment is 90% water. What is likely to happen?
A. water will rush into the cell
B. net movement of water will be equal
C. water will not move into or out of the cell
D. water will rush out of the cell

Answers

Water will rush into the cell, because water likes everything to be equal. This process is called diffusion (osmosis).

Determine the differences In electronegativity ionic radius atomic radius and first ionization energy for oxygen and beryllium

Answers

Answer:

Electronegativity  = 1.87.

Ionic radius = 109 pm.

Atomic radius = -39 pm

First ionization energy = 410 kJ/mol

Explanation:

Hello!

In this case, since electronegativity, ionic radius, atomic radius and first ionization energy are periodic properties that have specific trends, we can summarize it by realizing that oxygen and beryllium belong the same period 2 and differ in group, 6A and 2A respectively.

In such a way, the required comparison is written below:

Electronegativity = 3.44 (oxygen) - 1.57 (beryllium) = 1.87.

Ionic radius = 140 pm (oxygen)- 31 pm (beryllium) = 109.

Atomic radius = 73 pm (oxygen) - 112 pm (beryllium) = -39 pm

First ionization energy = 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium) = 410 kJ/mol

It means that electronegativity, ionic radius and first ionization energy increases from left to right whereas the atomic radius from right to left.

Best regards!

Oxygen and Beryllium are the elements that pertain to the same period and different groups 6A and 2A.

Ionic radius, atomic radius, electronegativity and first ionization energy are periodic qualities that have selective trends.

Differences in the trends are:

Ionic radius = 109 pm

Atomic radius = -39 pm

Electronegativity = 1.87

First ionization energy = 410 kJ/mol

This can be explained as:

Ionic radius:

= 140 pm (oxygen)- 31 pm (beryllium)

= 109.

Atomic radius:

= 73 pm (oxygen) - 112 pm (beryllium)

= -39 pm

Electronegativity:

= 3.44 (oxygen) - 1.57 (beryllium)

= 1.87

First ionization energy:

= 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium)

= 410 kJ/mol

Therefore, these data tell that ionic radius, electronegativity and first ionization energy increases from left to right in a period whereas, the atomic radius increases from right to left in a period.

To learn more about periodic trends follow the link:

https://brainly.com/question/12074167



If 8.793 of heat are added to a 2.00g sample of metal and the temperature increases by 4.9°C. What is the specific heat?

Answers

The specific heat of metal = 0.897 J/g° C  

Further explanation

Heat absorbed can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q = 8.793 J

m = 2 g

∆t = 4.9 °C

[tex]\tt c=\dfrac{Q}{m.\Delta t}\\\\c=\dfrac{8.793}{2\times 4.9}\\\\c=0.897~J/g^oC[/tex]

Calculate the mass of Na2O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction.

4NA + O2 ⟶ 2Na2O

Answers

Answer:

The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.

Explanation:

You know  the balanced reaction:

4 NA + O₂ ⟶ 2 Na₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:

Na: 4 molesO₂: 1 moleNa₂O: 2 moles

Being:

Na: 23 g/oleO: 16 g/mole

the molar mass of the compounds participating in the reaction is:

Na: 23 g/moleO₂: 2*16 g/mole= 32 g/moleNa₂O: 2*23 g/mole +16 g/mole=  62 g/mole

Then by stoichiometry of the reaction they react and are produced:

Na: 4 moles* 23 g/mole= 92 gO₂: 1 mole*32  g/mole= 32 gNa₂O: 2 moles* 62 g/mole= 124 g

Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?

[tex]massofNa_{2}O =\frac{4 grams of Na*124 gramsofNa_{2}O }{92 grams of Na}[/tex]

mass of Na₂O=5.39 g

The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.

What is the acceleration of gravity on Saturn if your mass is 75 kg and your weight is 800n

Answers

Answer:

The answer is 10.67 m/s²

Explanation:

The acceleration of gravity of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

f is the force

m is the mass

From the question we have

[tex]a = \frac{800}{75} = \frac{32}{3} \\ = 10.666666...[/tex]

We have the final answer as

10.67 m/s²

Hope this helps you

Calculate the total Calories in one cup of watermelon containing 0.9 g protein, 0.2 g fat, and 11 g carbohydrat

Answers

Answer:

49.4 cal

Explanation:

Step 1: Given data

Composition of 1 cup of watermelon: 0.9 g protein, 0.2 g fat, and 11 g carbohydrate.

Step 2: Calculate the calories provided by 0.9 g of protein

1 g of protein yields 4 cal.

0.9 g × (4 cal/1 g) = 3.6 cal

Step 3: Calculate the calories provided by 0.2 g of fat

1 g of fat yields 9 cal.

0.2 g × (9 cal/1 g) = 1.8 cal

Step 4: Calculate the calories provided by 11 g of carbohydrate

1 g of carbohydrate yields 4 cal.

11 g × (4 cal/1 g) = 44 cal

Step 5: Calculate the total number of calories

3.6 cal + 1.8 cal + 44 cal = 49.4 cal

3 moles of hydrogen occupy a volume of 67.20 L, at standard temperature and pressure (STP, 0 ˚C and 760 mmHg). What is the density of hydrogen at STP?

Answers

Answer:

The correct answer is 0.089 g/L ≅ 0.09 g/L

Explanation:

Density is defined as: mass/volume.

From the problem we have:

number of moles = n = 3 mol

volume = V = 67.20 L

We have to calculate the mass. For this, we need the molar mass (MM) of the gas. That is easily calculated from the molar mass of the element hydrogen (H), as we know that hydrogen gas has the molecular formula H₂:

MM(H₂) = 2 x molar mass H = 2 x 1 g/mol = 2 g/mol

Now, we multiply n by MM to obtain the mass (m) of the gas:

m = n x MM(H₂) = 3 mol x 2 g/mol = 6 g

Finally, we calculate the density from the mass and volume:

density = m/V = 6 g/(67.20) = 0.089 g/L ≅ 0.09 g/L

PLZZ ANSWER QUICKLY: True or False: MOST organelles found in plant cells are DIFFERENT than organelles found in animal cells

Answers

Answer:

Organelles allow for various functions to occur in the cell at the same time. Despite their fundamental similarities, there are some striking differences between animal and plant cells (see Figure 1). Animal cells have centrosomes (or a pair of centrioles), and lysosomes, whereas plant cells do not.

Explanation:

If I have an unknown quantity of gas at STP with a volume of 41 liters, how many moles of gas do
have?

Answers

Answer:1 mol

Explanation:

the conversion from mol to volume of gas at STP is 22.4L, meaning that the amount of mols in a gas of volume 22.4L at STP is one mol.

(7th grader sorry if its wrong)

Can a closed system gain particles

Answers

No because I had this question and got it right

which of the following requires the most energy to break a bond? a. breaking cl-br bond. b. breaking a n-p bond. c. breaking a o-o bond. d. none of the above.

Answers

A. Breaking a Cl-Br bond

A gas with a volume of 525 mL at a temperature of -25°C is heated to 175°C.

What is the new volume, in milliliters, of the gas if pressure and number of moles

are held constant?

Answers

Answer:

Volume V2 = 948.13 ml

Explanation:

Given:

Volume V1 = 525 ml

Temperature T1 = -25°C + 273.15

Volume V2 = ?

Temperature T1 = 175°C + 273.15

Computation:

V1 / T1 = V2 / T2

525 / [-25°C + 273.15] = V2 / [175°C + 273.15]

Volume V2 = 948.13 ml

The new volume, in milliliters, of the gas is 948 L

From the question,

We are to determine the new volume of the gas.

From Charles' law which states that the volume of a fixed mass of gas is directly proportional to the temperature (in Kelvin) provided that the pressure remains constant.

That is,

V ∝ T

Then,

V = kT

Therefore, we can write that

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

Where [tex]V_{1}[/tex] is the initial volume

[tex]T_{1}[/tex] is the initial temperature

[tex]V_{2}[/tex] is the final volume

and [tex]T_{2}[/tex] is the final temperature

From the given information,

[tex]V_{1} = 525 \ mL[/tex]

[tex]T_{1} = -25 ^{\circ}C = -25 + 273.15 \ K = 248.15 \ K[/tex]

[tex]T_{2} = 175 ^{\circ} C = 175 +273.15 \ K =448.15 \ K[/tex]

Putting the values into the formula, we get

[tex]\frac{525}{248.15} = \frac{V_{2} }{448.15}[/tex]

∴ [tex]V_{2} = \frac{525 \times 448.15}{248.15}[/tex]

[tex]V_{2} = \frac{235278.75}{248.15}[/tex]

[tex]V_{2} = 948.13 \ L[/tex]

V₂ ≅ 948 L

Hence, the new volume, in milliliters, of the gas is 948 L

Learn more here: https://brainly.com/question/12717895

According to the ideal gas law, a 10.68 mol sample of methane gas in a 0.8295 L container at 501.9 K should exert a pressure of 530.3 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure

Answers

Answer:

The answer is "152.28%".

Explanation:

Formula for ideal gas:

PV=nRT

equation:

[tex]\to (P+\frac{an^2}{v^2})(v-nb)=nRT \\\\a=2.253 \frac{L^2 atm}{mol^2}\\\\b=4.278 \times 10^{-2} \frac{L}{mol}\\\\n=10.68 \ mol\\\\v= 0.8295 \ L\\\\\to (P+\frac{an^2}{v^2})= \frac{nRT}{(v-nb)}\\\\\to (P)= \frac{nRT}{(v-nb)} - \frac{an^2}{v^2} \\\\[/tex]

          [tex]= \frac{10.68 \times 0.0082 \times 501.9}{0.8295 - (10.68 \times 4.278 \times 10^{-2})} - \frac{2.253 \times 10.68^2}{0.8295^2}\\\\=\frac{43.9543944}{0.4568904} - \frac{256.982587}{0.68807025}\\\\ = 96.2033661- 373.483067\\\\=-277.279701 \\\\ =-277.279701 \ atm[/tex]

The pressure of gas:

[tex]= (530.3 + 277.279701) \ atm\\\\= 807.579701 \ atm \\\\[/tex]

Calculating the pressure percentage:

[tex]=\frac{807.579701}{530.3} \times 100\\\\= 152.28 \%[/tex]

Chlorine dioxide is used as a disinfectant and bleaching agent. In water, it reacts to form chloric acid (HClO3), according to the following balanced equation: 6 ClO2 + 3 H2O → 5 HClO3 + HCl If excess ClO2 is mixed with 18.0 mL of H2O (d = 0.998g/1ml ) how many grams of HClO3 are formed?
would be very helpful if the answer was explained with steps

Answers

Answer:

b

Explanation:

b

Would someone mind helping me? I really need this answer but I'm so confused. I would appreciate any help :) and if you get the answer right, ill give you brainliest.

Answers

Answer:

A

Explanation:

liquids thake the shape of what holds them so shape is changing and volume is the same

your answer would be letter A!! goodluck with your quiz ig

2. A green laser pointer emits light at 554 nm. Calculate the frequency of this light and the energy of a photon of this light.

Answers

Answer:

Explanation:

Wavelength (λ) = speed/frequency

(λ) = c/f

Where;

c = speed (3× 10^8m/s)

f = frequency (Hz)

(

Calculate the molar solubility of mercury (I) bromide, Hg2Br2, in 1.0 M KBr. The Ksp for Hg2Br2 is 5.6 X 10−23. (Hint: How would the Br− concentration from the sparingly soluble compound itself compare to the Br− concentration that comes from the KBr?

Answers

Answer:

The correct answer is 5.6 × 10⁻²³ M.

Explanation:

As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.

Let the solubility of Hg₂Br₂ is S mol per liter.

Therefore,

KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)

Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)

Net [Br-] = (2S + 1) M

Ksp = S (2S + 1)²

Ksp = S (4S² + 1 + 4S)

Ksp = 4S³ + S + 4S²

As the solubility is extremely less, therefore, we can ignore S² and S³. Now,

Ksp = S = 5.6 × 10⁻²³ M

Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.

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