please help. calculate the partial pressure of oxygen gas Po2 at STP.

Answer:

0.140 Liter

Explanation:

M1V1 = M2V2

M1 = 1.61 M

V1 = 87 ml

M2 = 1.00 M

V2 = ?

V2 = M1V1/M2

V2 = 1.61×87/1

V2 = 140.07 ml

V2= 0.140 L

Answer:

D. -1882J

Explanation:

We can solve the energy released in a chemical reaction in an aqueous medium using the equation:

Q = -m*C*ΔT

Where Q is energy (In J),

m is mass of water (45.00g)

C is specific heat of water (4.184J/g°C)

And ΔT is change in temperature (25.00°C - 15.00°C = 10.00°C)

Replacing:

Q = -45.00*4.184J/g°C*10.00°C

Q = -1882J

Right answer is:

Answer:

You need to convert 27.8 miles to km and gallon to liters, then

(km/L) x #L = ?

27.8×112

=?

?×1gal

=?

?×1mi

=?

=?×?×?

answer

Answer:

Cs > Rb > K > Li

Explanation:

Given elements:

             Li,  Cs , K and Rb

Arrange in the the order of decreasing atomic radius:

Atomic radius is taken as half of the inter-nuclear distance (d) between two covalently bonded atoms of non - metallic elements or half of the distance between two nuclei in the solid state of metals.

So, the given species are the same group on the periodic table. These are group 1 elements;

          Li,  Cs , K and Rb

                   Cs > Rb > K > Li

Answer:

dependent

controlled

controlled

independent

independent

controlled

dependent

controlled

dependent

dependent

Explanation:

I HOPE IT HELPS :)

Answer:

0.92^n

Explanation:

Given that :

Initial amount of vinegar = 1 Litre

Number of litres removed repeatedly = 0.08 Litre

Since the amount removed each time is constant, then ;

Initial % = 100% = 100/100 = 1

. Using the relation :

Amount of vinegar in mixture :

Initial * (1 - amount removed / initial amount)^n

n = number of times repeated

1 * (1 - 0.08/1)^n

1 * (1 - 0.08)^n

1 * 0.92^n

Hence,

For nth removal,

Concentration will be :

0.92^n ; for n ≥ 1

Explanation:

Vanadium is in Period 4 and in the d-block of elements. The highest energy subshell is 4s. (because as electrons fill the 3d subshell, the subshell becomes of lower energy than 4s)

We have 1s²2s²2p⁶3s²3p⁶3d³4s².

Answer:

Vanadium (V) --- d-block element

Location on the periodic table :

Atomic number=23

Atomic mass=51 g/mol

The correct electron configuration for Vanadium is:

V= [Ar]4s²3d³ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³

3. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³ is the right answer.

Answer:

The sun

Explanation:

I did it and got it correct :-)

Answer:

is it c

Explanation:

Answer:

100 POINTS! (50 split up) PLSSSSSSSSSS HELP!!! I REALLY NEED IT SO MUCH :( I PROMICE TO GIVE BRAINLIEST AND FIVE STARS AND A THANKS! 6TH GRADE SCIENCE! i used all my points for this!

You measure the mass of an apple using a balance. You get these three measurements:

Trial 1: 167.0 g

Trial 2: 166.9 g

Trial 3: 167.2 g

You know that the true mass of the apple is 172 g. Describe your results in terms of precision and accuracy. Explain your answer in a paragraph for the best grade.

Explanation:

Answer:

ΔE =  -22 KJ

Explanation:

Given data:

Work done = 13 KJ

Heat lose = 9 KJ (exothermic process)

ΔE = ?

Solution:

Formula:

ΔE = q + w

ΔE = - 9KJ + (-13 KJ)   (work is done by the system on surrounding that's why we add negative sign)

ΔE =  -22 KJ

Answer

Two different elements have similar chemical properties when they have the same number of valence electrons in their outermost energy level. ... Elements in the same column of the Periodic Table have the same number of valence electrons – that's why they have similar chemical properties.

Statement that can be considered true as regards periodic table is : A: Elements in the same column share similar properties.

Therefore, option A is correct.

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Molar concentration of acetic acid = 1 M

Given

acetic acid in 6% solution (d=1 g/ml)

Required

molar concentration

Solution

mass of acetic acid in solution

mass = density x %mass

mass = 1 g/ml x 6%

mass = 0.06 g/ml ⇒ 60 g/L

MW acetic acid = 60 g/mol

molarity acetic acid = mass : MW

molarity = 60 g/L : 60 g/mol

molarity = 1 mol / L = 1 M

Answer : The mass in grams of calcium sulfate is 0.16 grams.

Explanation :

Molarity : It is defined as the number of moles of solute present in one litre of solution.

Formula used :

[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution}}[/tex]

Solute is, [tex]CaSO_4[/tex]

Given:

Molarity of [tex]CaSO_4[/tex] = 0.0025 mol/L

Molar mass of [tex]CaSO_4[/tex] = 136 g/mole

Volume of solution = 485 mL

Now put all the given values in the above formula, we get:

[tex]0.0025=\frac{\text{Mass of }CaSO_4\times 1000}{136\times 485}[/tex]

[tex]\text{Mass of }CaSO_4=0.16g[/tex]

Thus, the mass in grams of calcium sulfate is 0.16 grams.

Answer:

2nd one

Explanation:

Answer:

mass = 6.9 kg = 6900 g

volune= 722 cm3

density= mass/volume

density= 6900/722

= 9.55 g/ cm3

Answer:

[tex]\boxed {\boxed {\sf d \approx 9.56 \ g/cm^3}}[/tex]

Explanation:

We want to find the density of the plastic in grams per cubic centimeter.

Density is found by dividing the mass by the volume.

[tex]d=\frac {m}{v}[/tex]

We know the mass is 6.9 kilograms and the volume is 722 cubic centimeters We must convert kilograms to grams.

Now we know the mass and volume. Substitute the values into the density formula.  

[tex]d=\frac{ 6900 \ g }{722 \ cm^3}[/tex]

[tex]d= 9.5567867 \ g/cm^3[/tex]

Let's round to the nearest hundredth. The 6 in the thousandth place tells us to round the 5 to a 6.

[tex]d \approx 9.56 \ g/cm^3[/tex]

The density of the plastic is about 9.56 grams per cubic centimeter

Answer:

6

Explanation:

there are 2 sodium atoms, 1 carbon atom and 3 oxygen atoms

2+1+3=6 atoms

Answer:

seems like you got the answer already sooooooooooooooooooo

Explanation:

<3

Have a pic of Chenle dont hate

Answer:

A =8

B =8

C = O oxygen

D =26

E =30

F = Fe iron

Explanation:

No of protons = atomic number

No of protons + number of neutrons = atomic mass

Answer:

Using the periodic table, complete the table to describe each atom. Type in your answers.

A     ⇒ 20

B     ⇒ 40

C      ⇒ Ca

D     ⇒ 10

E      ⇒ 9

F      ⇒ F

Explanation:

yes

Answer:

B

Explanation:

No explanation needed for it i got 100% on it

Lifting the remains is the last step in the excavation of the skeletal remains. Therefore, option E is correct.

In archaeology, excavation can be described as the exposure, processing, and recording of archaeological remains. An excavation site or "dig" can be the area being studied. These locations are from one to various areas at a time while a project can be conducted over weeks to several years.

Excavation involves the recovery of various types of data from a site. This data includes ecofacts, artifacts, features, and archaeological context. Before excavating, the presence of remains can often be suggested by remote sensings, such as ground-penetrating radar.

During excavation, stratigraphic excavation removes phases of the site one layer at a time and keeps the timeline of the material remains consistent. This is done through mechanical means and the soil is processed via methods like mechanical sieving or water flotation.

Learn more about excavation of the skeletal remains, here:

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#SPJ2

Answer:

When the solar wind gets past the magnetic field and travels towards the Earth, it runs into the atmosphere .As the protons and electrons from the solar wind hit the particles in the Earth's atmosphere, they release energy – and this is what causes the northern lights.

Explanation:

Answer:

) B

Explanation:

Answer: There are [tex]1.08 \times 10^{-14}[/tex] moles [tex]N_{2}O_{4}[/tex] present in [tex]6.49 \times 10^9[/tex] particles.

Explanation:

According to the mole concept, 1 mole of a substance contains [tex]6.022 \times 10^{23}[/tex] particles.

Hence, number of moles present in [tex]6.49 \times 10^9[/tex] particles are calculated as follows.

No. of moles = [tex]\frac{6.49 \times 10^9}{6.022 \times 10^{23}}[/tex] mol

                      = [tex]1.08 \times 10^{-14}[/tex] mol

Thus, there are [tex]1.08 \times 10^{-14}[/tex] moles [tex]N_{2}O_{4}[/tex] present in [tex]6.49 \times 10^9[/tex] particles.

Answer:

1.08 × 10⁻¹⁴ moles of N2O4

Explanation:

Given data:

Moles of dinitrogen tetroxide = ?

Number of particles = 6.49×10⁹

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions , particles and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole contain 6.022 × 10²³ particles.

6.49×10⁹ particles × 1 mole / 6.022 × 10²³ particles

1.08 × 10⁻¹⁴ mol

Answer:

264.28 L of CO₂

Explanation:

We'll begin by calculating the number of moles present in 275 g of NO₃¯. This can be obtained as follow:

Mass of NO₃¯ = 275 g

Molar mass of NO₃¯ = 14 + (16×3) = 14 + 48 = 62 g/mol

Mole of NO₃¯ =?

Mole = mass / molar mass

Mole of NO₃¯ = 275 / 62

Mole of NO₃¯ = 4.44 moles

Next, we shall determine the number of mole of CO₂ produced from the reaction. This can be obtained as follow:

2NO₃¯ + 5CO + 2H⁺ —> N₂ + H₂O + 5CO₂

From the balanced equation above,

2 moles of NO₃¯ reacted to produced 5 moles of CO₂.

Therefore, 4.44 moles of NO₃¯ will react to produce = (4.44 × 5) / 2 = 11.1 moles of CO₂.

Thus, 11.1 moles of CO₂ were produced from the reaction.

Finally, we shall determine the volume of CO₂ produced each day. This can be obtained as follow:

Temperature (T) = 17 °C = 17 °C + 273 = 290 K

Pressure (P) = 1 atm

Number of mole (n) of CO₂ = 11.1 moles

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of CO₂ =?

PV = nRT

1 × V = 11.1 × 0.0821 × 290

V = 264.28 L

Thus, 264.28 L of CO₂ would be produced each day.

The volume of [tex]\rm CO_2[/tex] produced has been 264.28 L.

The volume of the carbon dioxide gas produced can be calculated with the ideal gas equation.

The moles of [tex]\rm NO_3^-[/tex] has been:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of 275 g  [tex]\rm NO_3^-[/tex] has been:

Moles = [tex]\rm \dfrac{275}{62}[/tex]

Moles of  [tex]\rm NO_3^-[/tex] = 4.44 moles

The balanced equation gives that with the reaction of 2 moles [tex]\rm NO_3^-[/tex] , 5 moles carbon dioxide has been formed.

Moles of carbon dioxide formed with 4.44 mol [tex]\rm NO_3^-[/tex] has been:

2 moles [tex]\rm NO_3^-[/tex] = 5 moles [tex]\rm CO_2[/tex]

4.44 moles [tex]\rm NO_3^-[/tex]  = 11.1 moles [tex]\rm CO_2[/tex].

The volume of the [tex]\rm CO_2[/tex] can be calculated as:

Pressure [tex]\times[/tex] Volume = moles [tex]\times[/tex] Rydberg constant [tex]\times[/tex] temperature

For the [tex]\rm CO_2[/tex] produced:

1 atm [tex]\times[/tex] Volume = 11.1 mol [tex]\times[/tex] 0.0821 atm.L/K.mol [tex]\times[/tex] 290 K

Volume = 264.28 L.

The volume of [tex]\rm CO_2[/tex] produced has been 264.28 L.

For more information about the volume of gas, refer to the link:

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The coefficient of Oxygen in the balanced reaction : 3

Given

Reaction(unbalanced)

C2H4+ O2--> CO2+H2O​

Required

The coefficient of Oxygen

Solution

C₂H₄+ aO₂⇒ bCO₂+cH₂O​

C, left=2, right=b⇒b=2

H, left=4, right=2c⇒2c=4⇒c=2

O, left=2a, right=2b+c⇒2a=2b+c⇒2a=2.2+2⇒2a=6⇒a=3

The equation becomes :

C₂H₄+ 3O₂⇒ 2CO₂+2H₂O​

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

From the question we have

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

From the question we have

[tex]V_2 = \frac{25 \times 30}{55} = \frac{750}{55} \\ = 13.636363...[/tex]

We have the final answer as

Hope this helps you

Answer:

P= 7.01 atm

P(CO₂)= 2.34 atm

Explanation:

Step 1: Convert the temperature to Kelvin

We will use the following expression.

K = °C + 273.15

K = 32.0°C + 273.15 = 305.2 K

Step 2: Calculate the total number of moles of the mixture

We will use the following expression.

n = nCO₂ + nN₂ = 2.33 mol + 4.66 mol = 6.99 mol

Step 3: Calculate the total pressure of the mixture

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T / V

P = 6.99 mol × 0.0821 atm.L/mol.K × 305.2 K / 25.0 L

P= 7.01 atm

Step 4: Calculate the partial pressure of carbon dioxide

We will use the ideal gas equation.

P(CO₂) × V = nCO₂ × R × T

P(CO₂) = nCO₂ × R × T / V

P(CO₂) = 2.33 mol × 0.0821 atm.L/mol.K × 305.2 K / 25.0 L

P(CO₂)= 2.34 atm