in order to balance the reaction, what number needs to go in front of F(2)?​

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU +  w

q = 475 J  +  93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

5) B

6) C

7) B newland, law of octaves

8) A)

Answer:

HNO3

Explanation:

i have the test

Therefore, decrease in PH can be achieved by adding acidic substance.

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Answer: There are 1.51 moles present in [tex]9.12 \times 10^{23}[/tex] atoms.

Explanation:

According to the mole concept, there are [tex]6.022 \times 10^{23}[/tex] atoms present in one mole of every substance.

Hence, moles present in [tex]9.12 \times 10^{23}[/tex] atoms are calculated as follows.

[tex]Moles = \frac{9.12 \times 10^{23}}{6.022 \times 10^{23}}\\= 1.51 mol[/tex]

Thus, we can conclude that there are 1.51 moles present in [tex]9.12 \times 10^{23}[/tex] atoms.

Answer:

I did it in steps in the picture I just hope it helps you to understand:)

ig The right answer is -Give the reaction more time.

Answer:

I think the answer is boiling

Answer:

No because it's located inside the nucleus of the atom

hope this helps

have a good day :)

Explanation:

An alpha particle emission causes the atomic number of a nuclide to decrease by 2 and its mass number to decrease by 4.

Nuclear Emission explains the release of energy stored in the nucleus of an atom in the form of particles like alpha particles, beta particles, gamma ray etc.

When a nucleus emits an alpha particle, it loses 2 protons and 2 neutrons, so its atomic number decreases by 2 and its mass number decreases by 4.

Hence, option 1 is correct.

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Answer:

Negative 5

Explanation:

I am not 100% sure btw

Answer:

2H2 in products

Explanation:

she should have added another 2 in H2

The edited equation does not balance the number of hydrogens. The correct balanced equation for the given reaction is written as follows:

[tex]\rm 2H_{2}O \rightarrow 2H_{2} +O_{2}[/tex].

The balanced chemical equation of a reaction represents the perfect stochiometric coefficients of all the reactants and products. Thus, each elements in the reaction have to be equal in both side.

According the mass conservation law, the total mass of the reactants is equal to the total mass of the products. Thus, all elements have to balanced on each side.

When water dissociates into hydrogen and oxygen molecules, 2moles of water undergoes dissociation to give two moles of H2 and one mole of O2. Thus the balanced equation is:

[tex]\rm 2H_{2}O \rightarrow 2H_{2} +O_{2}[/tex]

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The coefficients in a balanced equation can be used as molar ratios, which can act as conversion factors to relate the reactants to the products. These conversion factors state the ratio of reactants that react but do not tell exactly how much of each substance is actually involved in the reaction.

Answer:

The final volume is 6.4 L.

Explanation:

Dilution is reducing the concentration of a chemical and is achieved by adding more solvent to the same amount of solute. In other words, dilution is the procedure that is followed to prepare a less concentrated solution from a more concentrated one, and it simply consists of adding more solvent.

In a dilution the amount of solute does not vary, and as only more volume (and mass) of solvent is being added, the concentration of the solute decreases.

In a dilution the expression is used:

Ci*Vi = Cf*Vf

where:

In this case:

Replacing:

8 M* 2 L= 2.5 M* Vf

Solving:

[tex]Vf=\frac{8 M* 2 L}{2.5 M}[/tex]

Vf=6.4 L

The final volume is 6.4 L.

Answer:

molar mass (molar weight)

Explanation:

Explanation:

moles= number of particle or atom / 6.02×10^23

so

[tex] \frac{1.1 \times {10}^{20 } }{6.02 \times {10}^{23} } [/tex]

gives u 1.83×10^-4 moles

hope it make sense :)

Answer:

1.7 * 10^-4 moles

Explanation:

According to Avogadro, one mole of a substance contains 6.02 * 10^23 number of atoms, molecules, ions etc.

Hence;

1 mole of Na contains 6.02 * 10^23 atoms

x moles of Na contains  1.0x10^20 atoms

x = 1.0x10^20 * 1/6.02 * 10^23

x = 1.7 * 10^-4 moles

Answer:

Explanation:

cells process rna and trna

converting food into energy

the regrowth of cells that have died through out the day

the transcription of dna to rna

hope this helped :)

Answer:

because of heat and pressure

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial pressure (P1) = 1atm

Initial temperature (T1) = 0°C = 0°C + 273 = 273K

Final temperature (T2) = 280°C = 280°C + 273 = 553K

Final pressure (P2) =...?

Step 2:

Determination of the new pressure of the gas.

Since the volume of the gas is constant, the following equation:

P1/T1 = P2/T2

will be used to obtain the pressure. This is illustrated below:

P1/T1 = P2/T2

1/273 = P2 / 553

Cross multiply

273x P2 = 553

Divide both side by 273

P2 = 553/273

P2 = 2.03atm

Therefore, the new pressure of the gas will be 2.03atm

Answer:

[tex]\boxed {\boxed {\sf P_2=2.03 \ atm}}[/tex]

Explanation:

We are concerned with the variables of temperature and pressure, so we use Gay-Lussac's Law, which states the temperature of a gas is directly proportional to the pressure. The formula is:

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

We know that the container of gas begins at standard temperature and pressure (STP). This is 1 atmosphere of pressure and 273 Kelvin.

[tex]\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{T_2}[/tex]

We know the gas is put into an oven at 280 degrees Celsius. We can convert this to Kelvin.

[tex]\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{553.15 \ K}[/tex]

We are solving for the new pressure, so we must isolate the variable P₂. It is being divided by 553.15 Kelvin. The inverse of division is multiplication, so we multiply both sides by 553.15 K

[tex]553.15 \ K *\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{553.15 \ K} * 553.15 \ K[/tex]

[tex]553.15 \ K *\frac { 1 \ atm}{ 273 \ K}= P_2[/tex]

The units of Kelvin cancel.

[tex]553.15 *\frac { 1 \ atm}{ 273 }= P_2[/tex]

[tex]2.02619047619 \ atm = P_2[/tex]

Rounded to the nearest hundredth:

[tex]2.03 \ atm \approx P_2[/tex]

The new pressure is approximately 2.03 atmospheres.

Answer:

FeCl3(aq) -

3NaOH(aq) 0 Fe(OH)3(s) -

3NaCl(aq)

Explanation:

FeCl3(aq) -

3NaOH(aq) 0 Fe(OH)3(s) -

3NaCl(aq)

Answer:

Br

Explanation:

Answer:

It is a halogen element

Explanation:

hope this helps

have a great day/night

Answer:

[tex]\boxed {\boxed {\sf 2. \ 157 \ kPa}}[/tex]

Explanation:

Since temperature remains constant, the only variables that change are volume and pressure. Therefore, we are using Boyle's Law. This states that the pressure is inversely proportional to the volume. The formula is:

[tex]P_1V_1=P_2V_2[/tex]

We know the gas starts with a volume of 20.0 liters at a pressure of 275 kPa. We can substitute these values into the left side of the formula.

[tex]275 \ kPa *20.0 \ L=P_2V_2[/tex]

We know the gas expands to a volume of 35.0 Liters, but we do not know the pressure.

[tex]275 \ kPa *20.0 \ L=P_2* 35.0 \ L[/tex]

Since we are solving for the new pressure, we must isolate the variable P₂. It is being multiplied by 35.0 Liters and the inverse of multiplication is division. Divide both sides by 35.0 L.

[tex]\frac {275 \ kPa * 20.0 \ L}{ 35.0 \ L}= \frac{P_2*35.0 \ L}{35.0 \ L}[/tex]

[tex]\frac {275 \ kPa * 20.0 \ L}{ 35.0 \ L}=P_2[/tex]

The units of liters cancel.

[tex]\frac {275 \ kPa * 20.0 }{ 35.0 }=P_2[/tex]

[tex]\frac {5500}{35.0} \ kPa= P_2[/tex]

[tex]157.142857 \ kPa=P_2[/tex]

The original measurements of pressure and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

The 1 in the tenths place (157.142857) tells us to leave the 7 in the ones place.

[tex]157 \ kPa= P_2[/tex]

If the gas expanded to a volume of 35.0 liters while the temperature remained constant, the pressure in container was approximately 157 kilopascals.

Explanation:

A geological process where materials of Earth are transported from one place to another by natural forces like wind or water is called erosion.

When materials of Earth are deposited over a certain landmark due to transportation by natural forces then this process is called deposition.

The four major causes of erosion and deposition are as follows.

Gravity - Force that is responsible for attracting an object towards the center of Earth.

Running water - Water present in the rivers, oceans or lakes is moving from one place to another. Also, along with it carries pebbles or other materials.

Wind Waves - This is basically air and it keeps on moving from one place to another all the time. Wind waves tend to carry dust particles or other materials along with it.

Glaciers - When a mass of ice that is formed by accumulation and compaction of snow over mountains or around the poles tends to move slowly is called glaciers.

Answer: Electrolysis

Explanation: ‍♀️

Answer:

They equal the same

Explanation:

Because A ton of feathers and A ton of bricks. They both equal A ton.

Answer:

(C) pollen from flowers is blown by the wind.

I think this is the correct answer friend

Answer:

The required volume is 0.84 L

Explanation:

Given,

Mass of O2 gas = 1.20 g

We know,

At STP, molar volume of all gases is 22.4 L

and Molar mass of O2 gas is 32 g

Now,

Number of moles = mass/molar mass

                              = 1.20/32

                              = 0.0375

Again,

At STP,

Number of moles = Volume/Molar volume

or, Volume = Number of moles X Molar Volume

or Volume = 0.0375 X 22.4

∴ Volume = 0.84 L

So, the required volume is 0.84 L

Answer:

First Choice

it produces offspring with a large amount of genetic variation

Answer: i think this is correct.

Explanation: it occurs in simple organisms, such as bacteria, but not in multicellular organism such as plants and animals