If a car travels 400.0 meters in 20.0 seconds, how fast is it going? m/s Please help me

The chemical equation below shows the reaction of methane (CH4) and oxygen gas (O2) to form carbon dioxide (CO2) and water (H2O)

Leona [35]

The answer you have marked is correct

6 0

3 years ago

Fe+2HCLwhat the reaction type

S_A_V [24]

Answer:

FeCl2 + H. 2

Explanation:

Iron react with hydrogen chloride to produce iron(II) chloride and hydrogen.

7 0

3 years ago

If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml

Bumek [7]

Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

$\frac{V}{T} =k$

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

$\frac{P}{T} =k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T} =k$

Having two different states, an initial state and an final state, it is true:

$\frac{P1*V1}{T1} =\frac{P2*V2}{T2}$

In this case:

P1= 0.9 atm
V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

P2= 1 atm
V2= ?
T2= 0°C= 273 °K

Replacing:

$\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}$

Solving:

$V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}$

V2= 2.415 L =2,415 mL

The new volume is 2415 mL

6 0

3 years ago

The equilibrium constant of a reaction: A) is not related to the change in free energy of the reaction. B) is the same as the ma

Taya2010 [7]

Answer:) is related to the change in free energy of the reaction--d

Explanation:

For any reaction that is taking place at any moment the change in Gibbs Free Energy is related to the reaction quotient as

ΔG=ΔG⁰+RTlnQ

where R-Universal Gas Constant, T- Temperature in Kelvin, Q is the reaction quotient

Now when the system is in equilibrum, ΔG⁰ which is the standard Gibb's Free Energy,is then defined as

ΔG⁰=−RTlnK ,

where K is the equilibrium constant. because ΔG becomes 0 and reaction quotient Q = K

The equilibrum constant is related to the change in free energy of the reaction.

because when ΔG is negative, the value of K is high which leads to a spontaneous. reaction

when ΔG is positive, the value of K is low, which leads to a spontaneous. reaction in the opposite direction.

5 0

4 years ago

Given the following balanced equation, determine the rate of reaction with respect to [SO3]. If the rate of O2 loss is 3.56 x 10

Olenka [21]

Answer:

7.12 × 10⁻³ M/s

Explanation:

Step 1: Write the balanced reaction

2 SO₂ + O₂ ⇒ 2 SO₃

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of O₂ to SO₃ is 1:2.

Step 3: Calculate the rate of formation of SO₃

The rate of loss of O₂ is 3.56 × 10⁻³ mol O₂/L.s. The rate of formation of SO₃ is:

3.56 × 10⁻³ mol O₂/L.s. × 2 mol SO₃/1 mol O₂ = 7.12 × 10⁻³ mol SO₃/L.s

7 0

3 years ago