Which individual elements make up the following common

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76

aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.

The no. of moles can be determined by using the formula,

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.

4 0

3 years ago

A liquid has a _____.

guajiro [1.7K]

Answer:

definite volume

Explanation:

hope this helps!❆

3 0

2 years ago

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A 256 mL sample of HCl gas is in a flask where it exerts a force (pressure) of 67.5 mmHg. What is the pressure of the gas if it

Effectus [21]

Answer:

The pressure in the new flask would be $128\; \rm mmHg$ if the $\rm HCl$ here acts like an ideal gas.

Explanation:

Assume that the $\rm HCl$ sample here acts like an ideal gas. By Boyle's Law, the pressure $P$ of the gas should be inversely proportional to its volume $V$ .

For example, let the initial volume and pressure of the sample be $V_1$ and $P_1$ . The new volume $V_2$ and pressure $P_2$ of this sample shall satisfy the equation: $P_1 \cdot V_1 =P_2 \cdot V_2$ .

In this question,

The initial volume of the gas is $V_1= 256\; \rm mL$ .
The initial pressure of the gas is $P_1 = 67.5\; \rm mmHg$ .
The new volume of the gas is $V_2 = 135\; \rm mL$ .

The goal is to find the new pressure of this gas, $P_2$ .

Assume that this sample is indeed an ideal gas. Then the equation $P_1 \cdot V_1 =P_2 \cdot V_2$ should still hold. Rearrange the equation to separate the unknown, $P_2$ . Note: make sure that the units for $V_1$ and $V_2$ are the same before evaluating. That way, the unit of

$\begin{aligned} & P_2\\ &= \frac{P_1 \cdot V_1}{V_2} \\ &= \frac{256\; \rm mL \times 67.5\; \rm mmHg}{135\; \rm mL} \\ & \approx 128\; \rm mmHg\end{aligned}$ .

3 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago

Sample Response: Forces are either positive or negative vectors, depending on the direction in which the force is acting. The ne

lianna [129]

The net force is the sum of all of the forces that cancelled out to zero. Hence, option D is correct.

<h3>What is force?</h3>

The pushes or pulls on an object with mass causes it to change its velocity.

The net force is the vector sum of all forces acting on an object. Force is a vector quantity.

Two forces of equal magnitude when acting on an object in opposite directions will cancel out each other.

As a result net force is equal to 0. The word net here shows the total force acting on the object.

Hence, option D is correct.

Learn more about the force here:

brainly.com/question/14902131

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8 0

2 years ago

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