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Vlada [557]
3 years ago
13

A steel cylinder of oxygen is being store in a room at 25.0 °C under a pressure of 1250 atm. What pressure would be exerted by t

he gas if the container were transported across the desert at 40.0°C?
a. 1750 atm

b. 15700 atm

c. 1980 atm

d. 1310 atm
Chemistry
1 answer:
Rudik [331]3 years ago
6 0

Answer:

P₂ = 1312.88 atm

Explanation:

Given data:

Initial temperature = 25°C

Initial pressure = 1250 atm

Final temperature = 40°C

Final pressure = ?

Solution:

Initial temperature = 25°C (25+273.15 = 298.15 K)

Final temperature = 40°C ( 40+273.15 = 313.15 k)

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1250 atm / 298.15 K = P₂/313.15 K

P₂ = 1250 atm × 313.15 K / 298.15 K

P₂ = 391437.5 atm. K /298.15 K

P₂ = 1312.88 atm

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Answer:

a)M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88 g/mol

b) if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

Explanation:

0.2000 gHZ gives  100ml acid solution

33.5 ml of 0.1025 M NaOH  is required to prepare it

the moles = mass / molar mass

mass = 0.200 gHZ

moles = 0.0335*100 * 0.1025 = 0.034

therefore molar mass = mass / moles

                                  M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88

if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

3 0
3 years ago
What is the name of Bel on the periodic table
rewona [7]

Answer:

Nobelium or Beryllium

4 0
3 years ago
Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh
Aliun [14]

Answer:

1.60.

Explanation:

  • The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
  • The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>

<em></em>

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>

5 0
4 years ago
Give the symbol for the element with the following orbital diagram:
wel

Answer: 0%

Explanation:

You got a bad grade

5 0
3 years ago
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
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