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arlik [135]
2 years ago
15

A 3.0 L container holds a sample of hydrogen gas at 300 K and 101.5 kPa . The pressure increases to 305 kPa and the volume remai

ns constant. What will the temperature be?
Chemistry
1 answer:
Jlenok [28]2 years ago
8 0

Answer:

The temperature of the Hydrogen gas at pressure 305\ kPa is 901.4\ K.

Explanation:

Given the volume of the Hydrogen gas remain constant.

And the initial temperature and pressure are T_1=300\ K\ and\ P_1=101.5\ kPa

We need to find the temperature (T_2) of the gas at P_2=305\ kPa

We will use the gas equation to find out that temperature. As the volume is constant we can write with pressure and temperature.

\frac{P_1}{T_1}=\frac{P_2}{T_2}\\ \\T_2=\frac{P_2}{P_1}\times T_1\\\\T_2=\frac{305}{101.5}\times 300\\ \\T_2=3.004\times 300\\T_2=901.4\ K

The temperature of the Hydrogen gas at pressure 305\ kPa is 901.4\ K.

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By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm
kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

\tt 0.943~L=0.943\times 10^{-3}m^3

Volume of oil dumped = volume of swimming pool

\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}

3 0
2 years ago
A bag of fertilizer is labeled 10-20-20. What is the actual percentage of phosphorus in the fertilizer?
aleksandr82 [10.1K]

Answer:

20% of phosphorus

Explanation:

A fertilizer is used to improve the fertility of soils. Most fertilizers contains the element nitrogen, phosphorus and potassium.

They are often designated NPK fertilizers.

  Now we know that the numbers 10-20-20 depicts the nitrogen-phosphorus and potassium content of the fertilizer.

From the designation,

    The actual percentage is 20% of phosphorus.

                                               10%  of nitrogen

                                                20% of potassium

8 0
2 years ago
Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?
ioda

The correct answer is ClO, ClO3-, ClO- and ClO4-

Kossel and Lewis in 1916 developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electron in order to have an octet( 8 electron) in their shells. This is known as octet rule.

In ClO2-, oxygen contains 8 electrons in its valence shell and oxygen will share one electron with chlorine to complete the octet of Cl. In other four, we can clearly see that there are more or less than 8 electrons in the outer shell of oxygen so we can clearly say that ClO, ClO3-, ClO- and ClO4-  are disobeying the octet rule.

3 0
2 years ago
Read 2 more answers
Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta
Leokris [45]

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

3 0
2 years ago
The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in mil
jasenka [17]

Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Given data

Average distance between nitrogen and oxygen atoms: 115 pm

Step 2: Convert the distance to meters (SI base unit)

We will use the conversion factor 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Convert the distance to millimeters

We will use the conversion factor 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

5 0
2 years ago
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