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serious [3.7K]
3 years ago
13

Please help i have 2 min to complete

Chemistry
1 answer:
Lemur [1.5K]3 years ago
5 0
What’s the question
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A sample of a gas occupies 2.76 L at 303K. What would the volume be if the temperature was increased to 404K? Assume constant
Artist 52 [7]

Answer:

If the temperature was increased to 404 K, its volume would be 3.68 L.

Explanation:

Charles' Law gives a relationship between the volume and the temperature of the gas at constant temperature. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the temperature.

V\propto T

\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}

Let

V_1=2.76\ L\\\\T_1=303\ K\\\\T_2=404\ K

Let V_2 is new volume. Using above formula we get :

V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{2.76\times 404}{303}\\\\V_2=3.68\ L

If the temperature was increased to 404 K, its volume would be 3.68 L.

6 0
4 years ago
An electron in an atom is known to be in a state with magnetic quantum number . What is the smallest possible value of the princ
nlexa [21]

Answer:

Hi

Each electron in an atom is characterized by four numbers that arise from the resolution of Schrödinger's equations. These numbers are called quantum numbers. Each energy level corresponds to a main known quantum number, which is represented by the letter n. This number gives an idea of the location of an energy level with respect to the nucleus. The higher n, the mayor will be the energy of that level and the farther away from the nucleus is removed.

In each energy level there may be sub-levels. Each of them is specified by another quantum number called secondary, specified with the letter l. The value of this quantum number can vary from zero to n-1. For example, in the first energy level, the quantum number can only take a value that is zero, while in the second level, it can take a value between zero or one. Then, it can be said that the values of the quantum number n indicate the size of the orbital, that is, its proximity to the nucleus; and the values of the quantum number l variables the orbital:

• If l = 0, the orbital is of type s.

• If l = 1, the orbitals are of type p.

• If l = 2, the orbitals are of type d.

• If l = 3, the orbitals are of type f.

Explanation:

8 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
Note that 5° of longitude or latitude equals about 563 km. Use this information and a ruler to create a scale to use with this m
Ann [662]

Answer:

A suitable scale, say 1 cm: 100 km can be used.

Explanation:

Thinking process:

The best way to approach the question will be to consider the requirements. This is a simple case of scaling. In order to achieve the objective, you need to choose a scale that does not consume space and that presents more details at the same time.

For instance, a scale of 1 cm to 100 km will give me lines which are a little more then 5 cm. This can be presented as:

1: 500

This is appreciable for the paper size.

8 0
4 years ago
50 grams of acetic acid C2H4O2 are dissolved in 200 g of water. Calculate the weight % and mole fraction of the acetic acid in t
Len [333]
50g   of   C_2H_4O_2   in   200g    H_2O

so:

\frac{50}{200+50}+100\%= \frac{50*100}{250}\%=20\%

MW of acid = 2*C+4*H+2*O = 2*12+4*1+2*16=

=24+4+32=60g/mol

so:

\frac{50g}{60g/mol}\approx0.83moles

it means that

in 50g of acid there is \approx0.83moles of acid

MW of H_2O = 2*H+O=2*1+16=2+16=18g/mol

so:

\frac{200g}{18g/mol}\approx11.11moles

it means that:

in 200g of water there is \approx11.11moles of water

therefore:

\frac{0.83mol}{11.11mol+0.83mol}= \frac{0.83mol}{11.94mol}=0.069

So your answers are:

20\%

and the mole fraction is:

0.069
6 0
4 years ago
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