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Akito is warming a test tube over a Bunsen burner. He heats it too much and the glass breaks.Which best describes what Akito should do first?
Ο Ο Ο Ο
O get paper towels
O change lab stations
turn off the burner
remove his contacts

Answers

Answer 1
Answer: turn off the burner

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Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.1) Calculate q for the reaction. You must show your work.2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.3) Calculate ΔH for the reaction in kJ/mol. You must show your work.
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After a severe accident, a person can write and talk but has to learn towalk again. What part of the nervous system was probably affected? Explainwhy?

The solubility of glucose at 30°C is125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

Answers

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

400 mL * (0.996g)/(1mL) =398g

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

100gH_2O * (550gGlucose)/(398gH_2O) = 138 gGlucose

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated. If you want to increase the amount of glucose in the solution without adding more glucose, you can increase the temperature.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated.

Since the solubility of glucose at 30°C is 125 g/100 g water, adding 550 g of glucose to 400 mL of water exceeds the maximum amount of glucose that can dissolve in the given amount of water.

To increase the amount of glucose in the solution without adding more glucose, you can increase the temperature. Higher temperatures generally increase the solubility of solutes in water. By increasing the temperature, you can dissolve more glucose in the solution.

Learn more about solubility here:

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A lab group was calculating the speed of a radio car. They measured the distance traveled to be 6 meters and the time to be 3.5 seconds. Then they divided the distance by the time to find the speed. The actual speed was 2.2 m/s. What was their percent error?

Answers

A lab group was calculating the speed of a radio car. They measured the distance traveled to be 6 meters and the time to be 3.5 seconds. Then they divided the distance by the time to find the speed. The actual speed was 2.2 m/s. Their percent error is 22.1%.

Percent error is a measure of the difference between an observed value and a true value.

Actual Speed (True Value) = 2.2 m/s

Experimental Speed (Calculated Value) = Distance / Time = 6 m / 3.5 s = 1.714 m/s

The formula for calculating percent error is:

Percent Error = ((|Actual Value - Experimental Value|) / |Actual Value|) * 100%

Calculate the absolute difference between the actual speed and the experimental speed:

|2.2 - 1.714| = 0.486

Calculate the absolute value of the actual speed:

|2.2| = 2.2

Percent Error = (0.486 / 2.2) * 100%

= 0.221 * 100%

= 22.1%

The calculated percent error is approximately 22.1%. This means that the lab group's calculated speed of 1.714 m/s is about 22.1% lower than the true speed of 2.2 m/s.

Percent error is a way to quantify the accuracy of experimental measurements. A positive percent error indicates that the experimental value is higher than the true value, while a negative percent error indicates that the experimental value is lower. In this case, since the calculated speed is lower than the true speed, we have a positive percent error.

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Answer:456

Explanation:

Assuming 54.81 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be produced if the reaction will only produce 66.93 % yield?

Answers

Answer:

181.39g of AlCl3 is produced

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3CuCl2•2H2O + 2Al → 2AlCl3 + 6H2O + 3Cu

Next, we shall determine the mass of Al that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:

Molar mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Summary:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Next, we shall determine the theoretical yield of AlCl3. This can be achieved as shown below:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Therefore, 54.81g of Al will react to produce = (54.81 x 267)/54 = 271.01g of AlCl3.

Therefore, the theoretical yield of AlCl3 is 271.01g.

Finally, we shall determine the actual yield of AlCl3 produced from the reaction.

This can be obtain as follow:

Percentage yield of AlCl3 = 66.93%

Theoretical yield of AlCl3 = 271.01g

Actual yield of AlCl3 =?

Percentage yield = Actual yield/Theoretical yield x 100

66.93% = Actual yield /271.01g

Actual yield = 66.93% x 271.01

Actual yield = 66.93/100 x 271.01g

Actual yield = 181.39g.

Therefore, 181.39g of AlCl3 is produced from the reaction.

What is the molarity of the potassium hydroxide if 27.20 mL of KOH is required to neutralize 0.604 g of oxalic acid, H2C2O4?H2C2O4(aq)+2KOH(aq)→K2C2O4(aq)+2H2O(l)

Answers

Answer:

Concentration of KOH = 1.154 M

Explanation:

H_2C_2O_4(aq) + 2KOH(aq) \rightarrow K_2C_2O_4(aq) + 2H_2O(l)

In the above reaction, 1 mole of oxalic acid reacts with 2 moles of KOH.

Mass of oxalic acid = 0.604 g

Mole = (Mass\; in\;g)/(Molecular\;mass)

Molecular mass of oxalic acid = 90.03 g/mol

Mole = (0.604)/(90.03)=0.0067\;mol

1 mol of oxalic acid reacts with 2 moles of KOH

0.0067 mol of oxalic acid reacts with 0.0067* 2 = 0.0134 mol\; of\;KOH

Volume of the solution = 27.02 mL = 0.0272 L

Molarity=(Mole)/(Volume\;in\;L)

No. of mole of KOH = 0.0134 mol

Molarity=(0.0134)/(0.0272)=1.154\;M

Concentration of KOH = 1.154 M

Write a balanced nuclear equation for the following: The nuclide uranium-238 undergoes alpha emission.

Answers

Answer:

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He + Energy

Explanation:

Hello,

When uranium undergoes an alpha decay, it's atomic number is reduced by 2 while it atomic mass is reduced by 4 similar to the elements helium (He) and sometimes alpha decay is represented by a helium atom.

Equation of decay

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He + Energy

When uranium undergoes an alpha decay, it loses an alpha particle similar to a helium atom and forms thorium which has an atomic number of 90 and atomic mass of 234 with the release of energy.

The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 4.25 kJ·mol−1 at 2000. K and −63.12 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature.

Answers

Answer:

K^(2000K)=0.774\n\nK^(3000K)=12.56

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

\Delta _rG=\Delta _fG_(X)-(1)/(2) \Delta _fG_(X_2)=\Delta _fG_(X)

Thus, at 2000 K:

\Delta _rG=\Delta _fG_(X)^(2000K)=4.25kJ/mol

And at 3000 K:

\Delta _rG=\Delta _fG_(X)^(3000K)=-63.12kJ/mol

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

K=exp(-(\Delta _rG)/(RT) )

Thus, at each temperature we obtain:

K^(2000K)=exp(-(4250J/mol)/(8.314(J)/(mol* K)*2000K) )=0.774\n\nK^(3000K)=exp(-(-63120J/mol)/(8.314(J)/(mol* K)*3000K) )=12.56

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.
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