Al (s) + HCl (aq) → H2 (g) + AlCl3 (aq)

This is an example of:

A. Double replacement

B. Single replacement

C. Synthesis

D. Decomposition

Al (s) + HCl (aq) H2 (g) + AlCl3 (aq) This Is An Example Of:A. Double ReplacementB. Single ReplacementC.

Answers

Answer 1

Answer:

B. Single replacement


Related Questions

What volume of a 1.2M solution must be used to produce .5 L of a .7M solution?

Answers

Answer:

1,2million or meter

Explanation:

or 1 million until 7m

Consider a nonadiabatic well-stirred reactor with simplifi ed chemistry, i.e., fuel, oxidizer, and a single product species. the reactants, consisting of fuel (yf = 0.2) and oxidizer (yox = 0.8) at 298 k, fl ow into the 0.003-m3 reactor at 0.5 kg / s. the reactor operates at 1 atm and has a heat loss of 2000 w. assume the following simplifi ed thermodynamic properties: cp = 1100 j / kg-k (all species), mw = 29 kg / kmol (all species), hf f o , = −2000 kj/ kg, hf ox o , = 0, and hf o , pr = −4000 kj/ kg. the fuel and oxidizer mass fractions in the outlet stream are 0.001 and 0.003, respectively. determine the temperature in the reactor and the residence ti

Answers

The first step is to calculate the molar flow rate of fuel, oxidizer, and product. This is done by dividing the mass flow rate (0.5 kg/s) by the molecular weight of each species (29 kg/kmol).

What is molecular?

Molecular is a term used to describe the smallest units of matter. Molecules are made up of atoms and are held together by a chemical bond, which involves the sharing of electrons between atoms.

This gives us the following molar flow rates:

Fuel: 0.017 kmol/s

Oxidizer: 0.027 kmol/s

Product: 0.046 kmol/s

Next, we need to calculate the enthalpy change for the reaction. Since we are dealing with a single product species, the enthalpy change can be calculated using the following equation:

ΔH = (hf f o , + hf ox o , - hf o , pr) * n

Where:

hf f o , = Enthalpy of formation of fuel

hf ox o , = Enthalpy of formation of oxidizer

hf o , pr = Enthalpy of formation of product

n = Molar flow rate of product

Substituting the given values, we get the following:

ΔH = (-2000 + 0 - (-4000)) * 0.046 = 920 kJ/s

Now we can calculate the heat of reaction by multiplying the enthalpy change with the molar flow rate of the reactants. This gives us the following result:

Heat of reaction = (0.017 kmol/s * 920 kJ/s) + (0.027 kmol/s * 920 kJ/s) = 24.12 kJ/s

We can then calculate the temperature of the reactor by subtracting the heat loss (2000 W) from the heat of reaction and dividing by the total mass flow rate of the reactants (0.5 kg/s) multiplied by the specific heat capacity (1100 J/kg-K). This gives us the following result:

T = (24.12 kJ/s - 2000 W) / (0.5 kg/s * 1100 J/kg-K) = 436 K

Finally, we can calculate the residence time by dividing the volume of the reactor (0.003 m3) by the total mass flow rate of the reactants (0.5 kg/s). This gives us the following result:

Residence time = 0.003 m3 / 0.5 kg/s = 0.006 s

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The temperature in the reactor is approximately 10.74 K. and 0.006 s.  

The temperature in the reactor and the residence time, we need to solve the following set of equations:

dU = w + Q / m

Next, we need to find the rate of change of mass flow rate, which is given by:

dm = Fv - D

here Fv is the volume flow rate of reactants and D is the diffusion rate of the product.

Finally, we can use the above equations to find the temperature in the reactor and the residence time as follows:

Temperature in the reactor:

T = (dU / Q) / (m / cP)

here cP is the specific heat at constant pressure.

Residence time:

t = (m / D)

We can assume that the reactants have a volume flow rate of 0.5 kg/s and the product species has a volume flow rate of 0.001 kg/s. Therefore, the mass flow rate of the reactants is:

m = 0.5 kg/s * 0.002 m3/kg = 0.001 kg/s

The diffusion rate of the product can be calculated as:

D = k * (yox - yf) / (yf + yox)

here k is the reaction rate constant and (yox - yf) / (yf + yox) is the molar fraction of the product species.

Using the values of k, m, and (yox - yf) / (yf + yox) from the problem statement, we can calculate the diffusion rate of the product as:

D = 1 * (0.003 - 0.2) / (0.2 + 0.003)

= 0.00006 / 0.003

= 0.1833

Therefore, the residence time of the reactor is:

t = (0.001 kg/s / 0.1833 kg/mol) = 0.051 s

The temperature in the reactor is given by:

T = (dU / Q) / (m / cP)

here cP is the specific heat at constant pressure of the reactants, which is 1100 J/kg-K.

T = (w + Q / m) / (0.001 kg/s / 1100 J/kg-K) / (0.001 kg/s / 0.003 m3/kg)

= 10.74 K

Residence time = 0.003 m3 / 0.5 kg/s = 0.006 s

Therefore, the temperature in the reactor is approximately 10.74 K and 0.006 s.  

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35 POINTS -- REAL ANSWERS (please)



For each of your three trials state the following:



⢠heat needed to melt the ice (q) (I got 18* for all)


⢠enthalpy of fusion (I'm not sure how to find the mass of the ice melted)


⢠percent error from the accepted enthalpy of fusion of water of 334 J/g (I don't understand this, we never went over this)

Answers

To calculate the enthalpy of fusion and percent error for each of your three trials. Here are the steps to calculate each value:

1. Heat needed to melt the ice (q): You've already mentioned that you have this value as 18* for all three trials. I'm assuming this is in joules (J).

2. Enthalpy of fusion (ΔHfus): To calculate this, you need the mass of the ice melted (m). You mentioned that you're not sure how to find the mass of the ice melted. Usually, this value is provided in the experiment or you can measure it using a scale. Once you have the mass, use the following formula:

ΔHfus = q / m

3. Percent error: To calculate the percent error, you need the accepted enthalpy of fusion of water, which is 334 J/g. Use the following formula:

Percent error = (|calculated ΔHfus - accepted ΔHfus| / accepted ΔHfus) × 100

Now, perform these calculations for each of your three trials. Note that you'll need to obtain or measure the mass of the ice melted (m) for each trial to calculate the enthalpy of fusion and percent error.

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What is the pH of a solution that is 0. 17M HA and 0. 50M A-.  Ka HA=2. 87x10-9​

Answers

The pH of a solution that is 0.17M HA and 0.50M A- can be calculated using the Henderson-Hasselbalch equation. This equation states that the pH of a solution is equal to the pKa of the acid plus the log of the ratio of the conjugate base to the acid.

In this case, the pKa of HA is 2.87x10-9, and the ratio of A- to HA is 0.50/0.17 which is roughly 2.94. Therefore, the pH of this solution is 2.87x10-9 + log(2.94) = -6.53.

To arrive at this result, the equation takes into account the fact that HA is the acid and A- is the conjugate base. HA donates a proton to A- in aqueous solution, forming the HA- and A2- ions.

The ratio of A- to HA is a measure of the amount of protonation that has occurred, and the pKa is the pH at which the protonation is equal. The Henderson-Hasselbalch equation shows us how the ratio of conjugate base to acid affects this equilibrium, allowing us to calculate the pH of the solution.

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