Why does a leak in a car tire (at constant temperature) decrease the pressure of the tire?A) It removes kinetic energy from the system.
B) The number of molecules decreases.
C) The temperature decreases.
D) The volume decreases.

Answers

Answer 1

Answer:

Explanation:

The volume decreases.

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If the pressure on an ideal gas is increased, what will happen to the volume of the gas?

Answers

If the pressure on an ideal gas is increased, the volume of the gas will decrease. This can be predicted with the use of the ideal gas equation which is expressed as: PV = nRT. At constant temperature, we can say that pressure and volume are inversely related. Thus, as one value increase, the other decrease.

450

Explanation:

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Calculate the heat energy for this scenario: You are repeating the food experiment we did in class with a different food snack. At the beginning of this experiment, you started with 25 grams of water at 22 deg * C At the end of the experiment, the final temperature of the water is 45°C. The specific heat of water is 4.18J / (deg) * CSub and solve

Answers

Answer:

Mass of water (m) = 25 grams = 0.025 kg (since 1 g = 0.001 kg)

Specific heat of water (c) = 4.18 J/(g°C) = 4.18 J/(kg°C)

Initial temperature ( $T_ {initial}$ ) = 22°C

Final temperature ( $T_(final)$ )= 45°C

Change in temperature (ΔT):

ΔT= $T_(final)$ - $T_ {initial}$ =45°−22°=23°

Now, calculate the heat energy (Q)

Q=mass×specific heat×ΔT

Q=0.025kg×4.18J/(kg°C)×23°C

Q≈2.44kJ

So, the heat energy for this scenario is approximately 2.44 kilojoules (kJ).

Which will cause a decrease in gas pressure in a closed containerA) Reducing the volume
B) Adding more gas
C) Lowering the temperature

Answers

By lowering the temps, there is a decrease in the pressure inside. so C.

Answer: C, Lowering the temperature

Explanation: Gradpoint

Democritus was a Greek philosopher living from c. 460 B.C. to c. 370 B.C. He wrote several suppositions on the properties of matter.Which of the following statements most closely describes Democritus's ideas about matter?

A. All matter is composed of tiny particles that can vary in shape, size, and weight.

B. Solid matter is composed of square-shaped particles and liquid matter is composed of flat particles.

C. Solid matter is composed of atoms and liquid matter is composed of elements.

Answers

The correct answer is answer choice A, all matter is composed of tiny particles that can vary in shape, size, and weight. Democritus believed that the universe is made up of empty space and tiny particles; these particles were atoms.

Hope this helps!

If 4.00 g of NaCl react with 10.00 g of AgNO3, what is the excess reactant?

Answers

Answer : The excess reagent is, $NaCl$

Explanation :

First we have to calculate the moles of $NaCl$ and $AgNO_3$ .

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=(4g)/(58.4g/mole)=0.068moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=(10g)/(169.9g/mole)=0.058moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NaCl+AgNO_3\rightarrow AgCl+NaNO_3$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $AgNO_3$

So, 0.068 moles of $NaCl$ react with 0.068 moles of $AgNO_3$

From this we conclude that, the moles of $AgNO_3$ are less than the NaCl. So, $AgNO_3$ is a limiting reagent because it limits the formation of products and $NaCl$ is an excess reagent.

Hence, the excess reagent is, $NaCl$

The reaction formula of this is NaCl + AgNO3 = NaNO3 + AgCl. The mole number of NaCl is 4/58.5=0.068 mol. The mole number of AgNO3 is 10/170=0.059 mol. So the NaCl is excess.

What mass of magnesium bromide is formed when 1.00 g of magnesium reacts with 5.00 g of bromine?

Answers

Answer:

when 1.00 g of magnesium reacts with 5.00 g of bromine, approximately 7.57 g of magnesium bromide is formed.

Explanation:

To find the mass of magnesium bromide formed when 1.00 g of magnesium reacts with 5.00 g of bromine, you need to first write a balanced chemical equation for the reaction between magnesium and bromine. The balanced equation for the formation of magnesium bromide (MgBr2) is as follows:

Mg + Br2 → MgBr2

Now, you can calculate the molar mass of each substance involved in the reaction:

Molar mass of Mg (magnesium) = 24.31 g/mol

Molar mass of Br2 (bromine) = 2 * 79.90 g/mol = 159.80 g/mol

Molar mass of MgBr2 (magnesium bromide) = 24.31 g/mol + 2 * 79.90 g/mol = 184.11 g/mol

Next, calculate the number of moles for each reactant:

Moles of Mg = Mass (1.00 g) / Molar mass (24.31 g/mol) = 0.0411 moles

Moles of Br2 = Mass (5.00 g) / Molar mass (159.80 g/mol) = 0.0313 moles (approximately, rounded to four decimal places)

Now, determine the limiting reactant. To do this, compare the mole ratio between Mg and Br2 in the balanced equation. The balanced equation shows that 1 mole of Mg reacts with 1 mole of Br2. Therefore, the limiting reactant is the one that is present in the smaller amount relative to the balanced equation's stoichiometry.

In this case, magnesium (0.0411 moles) is present in a smaller amount than bromine (0.0313 moles). So, magnesium is the limiting reactant.

Now that you know magnesium is the limiting reactant, you can calculate the mass of magnesium bromide formed using the stoichiometry of the balanced equation. According to the balanced equation, 1 mole of Mg produces 1 mole of MgBr2.

Moles of MgBr2 formed = Moles of Mg (limiting reactant) = 0.0411 moles

Now, calculate the mass of magnesium bromide formed:

Mass of MgBr2 = Moles of MgBr2 × Molar mass of MgBr2

Mass of MgBr2 = 0.0411 moles × 184.11 g/mol = 7.57 g

So, when 1.00 g of magnesium reacts with 5.00 g of bromine, approximately 7.57 g of magnesium bromide is formed.

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