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How many grams of CO2 and H2O are produced from the combustion of 220. g of propane (C3H8)? C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Answers

Answer 1

Answer:

Answer:

(C3H8) produces 660 g of CO2 and 360 g of H2O

Explanation:

The balanced chemical equation for the combustion of propane (C3H8) is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

This equation tells us that for every molecule of propane (C3H8) that reacts with 5 molecules of oxygen (O2), 3 molecules of carbon dioxide (CO2) and 4 molecules of water (H2O) are produced.

So, if we have 220. g of propane (C3H8), we can find the amount of CO2 and H2O produced by using the mole ratio from the balanced equation:

1 mole C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

We can find the number of moles of C3H8 by dividing the mass by the molar mass of C3H8 (44 g/mol):

220 g / 44 g/mol = 5 moles C3H8

So, the number of moles of CO2 and H2O produced can be found by multiplying the number of moles of C3H8 by the mole ratio:

3 moles CO2 = 3 moles CO2/1 mole C3H8 * 5 moles C3H8 = 15 moles CO2

4 moles H2O = 4 moles H2O/1 mole C3H8 * 5 moles C3H8 = 20 moles H2O

Finally, we can convert the number of moles of CO2 and H2O to grams by multiplying by their molar masses (44 g/mol for CO2 and 18 g/mol for H2O):

15 moles CO2 * 44 g/mol = 660 g CO2

20 moles H2O * 18 g/mol = 360 g H2O

So, the combustion of 220 g of propane (C3H8) produces 660 g of CO2 and 360 g of H2O.

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How many moles are in 12.0 grams of O2

Answers

Answer:

Moles = 0.375

Explanation:

Moles= m/M

= 12/32 = 0.375mol

The genetic code from DNA is carried into the
cytoplasm by____

Answers

Answer:

The type of RNA that contains the information for making a protein is called messenger RNA (mRNA) because it carries the information, or message, from the DNA out of the nucleus into the cytoplasm.

Answer:

Messenger RNA (mRNA)

Explanation:

Messenger RNA (mRNA), Molecules in cells that carries codes from the DNA in the nucleus to the sites of protein synthesis in the cytoplasm (the ribosomes)...

Hope This helps if not Then im Sorry XD

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH . It requires 11.9 mL of the NaOH solution to reach the end point of the titration. A buret filled with a titrant is held above a graduated cylinder containing an analyte solution. What is the initial concentration of HCl

Answers

Answer:

0.190 M

Explanation:

Let's consider the neutralization reaction between HCl and NaOH.

HCl + NaOH = NaCl + H2O

11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:

0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol

The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.

1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:

M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M

Answer:

The initial concentration of HCl was 0.1904 M

Explanation:

Step 1: Data given

Volume of HCl solution = 10.0 mL = 0.010 L

Volume of a NaOH solution = 11.9 mL = 0.0119 L

Molarity of NaOH solution = 0.160 M

Step 2: The balanced equation

HCl + NaOH → NaCl + H2O

Step 3: Calculate the concentration of HCl

C1*V1 = C2*V2

⇒with C1 = the concentration HCl = TO BE DETERMINED

⇒with V1 = the volume of HCl = 0.010 L

⇒with C2 = the concentration of NaOH = 0.160 M

⇒with V2 = the volume of NaOH = 0.0119 L

C1 * 0.010 L = 0.160 M * 0.0119 L

C1 = (0.160 M * 0.0119 L) / 0.010 L

C1 = 0.1904 M

The initial concentration of HCl was 0.1904 M

A certain process has ΔH° > 0, ΔS° < 0, and ΔG° > 0. The values of ΔH° and ΔS° do not depend on the temperature. Which of the following is a correct conclusion about this process? None of the above conclusions is correct. It is non-spontaneous at all T. It is spontaneous at low T. It is spontaneous at all T. It is spontaneous at high T

Answers

Answer: It is non-spontaneous at all T.

Explanation:

According to Gibb's equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy = +ve

$\Delta H$ = enthalpy change = +ve

$\Delta S$ = entropy change = -ve

T = temperature in Kelvin

$\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

Putting in the values:

$\Delta G=(+ve)-T(-ve)$

$\Delta G=(+ve)(+ve)=+ve$

Reaction is non spontaneous at all temperatures.

The following data is given to you about a reaction you are studying: Overall reaction: 2A  D Proposed mechanism: Step 1 A + B  C (slow) Step 2 C + A  D + B (fast) [A]o = 0.500 M [B]o = 0.0500 M [C]o = 0.500 M [D]o = 1.50 M This reaction was run at a series of temperatures and it was found that a plot of ln(k) vs 1/T (K) gives a straight line with a slope of -982.7 and a Y intercept of -0.0726. What is the initial rate of the reaction at 298K?

Answers

Answer : The initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

Explanation :

The Arrhenius equation is written as:

$K=A* e^{(-Ea)/(RT)}$

Taking logarithm on both the sides, we get:

$\ln k=-(Ea)/(RT)+\ln A$ ............(1)

where,

k = rate constant

Ea = activation energy

T = temperature

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor

The equation (1) is of the form of, y = mx + c i.e, the equation of a straight line.

Thus, if we plot a graph of $\ln k$ vs $(1)/(T)$ then the graph shows a straight line with negative slope. That means,

Slope of the line = $-(Ea)/(R)$

And,

Intercept = $\ln A$

As we are given that:

Slope of the line = -982.7 = $-(Ea)/(R)$

Intercept = -0.0726 = $\ln A$

Now we have to calculate the value of rate constant by putting the value of slope, intercept and temperature (298K) in equation 1, we get:

$\ln k=-(982.7)/(298)+(-0.0726)$

$\ln k=-3.37$

$k=0.0344s^(-1)$

The value of rate constant is, $0.0344s^(-1)$

Now we have to calculate the initial rate of the reaction at 298 K.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightarrow C$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

As we are given that:

[A] = 0.500 M

[B] = 0.0500 M

k = $0.0344s^(-1)$

Now put all the given values in the rate law expression, we get:

$Rate=(0.0344)* (0.500)* (0.0500)$

$Rate= 8.6* 10^(-4)M/s$

Therefore, the initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

The chemical formula for potassium bromide is KBr.A chemist determined by measurements that 0.0250 moles of potassium bromide participate in a chemical reaction. Calculate the mass of potassium bromide that participates.

Round your answer to 3 significant digits.

Answers

A chemist determined by measurements that 0.0250 moles of potassium bromide participate in a chemical reaction is 0.002 moles

What are moles?

The moles are the smallest unit of an atom ion molecule or substance which is used to count the number which is taking part in a chemical reaction and is equal to 2.303 ×10²³ moles of that.

To calculate the participant in a chemical reaction number of moles is 0.0250 moles so the mass will be,

number of moles = mass/ molar mass

substituting the value,

0.0250 moles = mass / 119.002

mass = 119.002 × 0.0250 moles

mass = 0.002 moles

Therefore, 0.002 moles determined by measurements that 0.0250 moles of potassium bromide participate in a chemical reaction.

Learn more about moles, here:

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My solution

39.0983+126.90447=166.00277