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Tcecarenko [31]
2 years ago
6

Give evidence to support or dispute: “In nature, the chance of finding one isotope of an element is the same for all elements.”

Chemistry
1 answer:
alekssr [168]2 years ago
5 0

Answer:

False, isotopes have different occurrence percentages, so the changes are different.

Explanation:

Hello,

In this case, since it is false that the isotopes of all the elements can be found with the same chance (occurrence) we can consider the following facts:

1. Carbon atom has two major occurring isotopes: C-12 (98.93%) and C-13 (1.07%).

2. Bromine atom has two major occurring isotopes: Br-79 (50.69%) and Br-81 (49.31%).

3. Calcium has four major occurring isotopes: Ca-40 (96.941%), Ca-42 (0.647%), Ca-43 (0.135%) and Ca-44 (2.086%).

Which show us that the chances of finding any isotope differ among elements.

Regards.

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in collecting the precipitate, why would it be inappropriate to heat the reacted mixture and evaporate off the water?
djverab [1.8K]
In collecting the precipitate, it is inappropriate to heat <span>the reacted mixture and evaporate off the water because it is possible that the mixture contains other substances that precipitates as well when the mixture is being heated so you will not be able to collect what you want.</span>
6 0
2 years ago
Olive oil has a density of 0.92g/ml. How much would 1.0 Liter of olive oil weigh in grams?
Julli [10]

Answer:

9.2x10²g

Explanation:

Data obtained from the question include the following:

Density = 0.92g/ml

Volume = 1L = 1 x 1000 = 1000mL

Mass =..?

Density is simply defined as the mass of the substance per unit volume of the substance. Mathematically it can be represented as:

Density = Mass /volume.

Mass = Density x volume

Mass = 0.92 x 1000

Mass = 9.2x10²g.

Therefore, 1L of olive will weigh 9.2x10²g.

5 0
2 years ago
________ is the study of matter and the energy that causes matter to combine, break apart and recombine in everything living and
Kipish [7]

Answer: Emperical formula

Explanation:

8 0
2 years ago
BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: K_{b} = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: K_{w} = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
2 years ago
495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F
I am Lyosha [343]

Answer:

<u><em>Total pressure of the flask is 2.8999 atm.</em></u>

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

                                              = 0.495 L (1 cm³ = 1 mL = 0.001 L)                                            

Volume of nitrogen (N2) gas =  877 cm3

                                               = 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

                         = 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature =  25 °C

T = (25°C + 273.15) K

    = 298.15 K

Pressure = 114.7 kPa

               = 114.700 Pa

Pressure (torr) = 114,700 / 101325

                        = 1.132 atm

Formula:

PV=nRT  <em>(ideal gas equation)</em>

P = pressure

V = volume

R (gas constnt)=  0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

<u>For Oxygen:</u>

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

  = 0.560 / 24.47

  = 0.0229 moles

<u>For Nitrogen:</u>

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

  = 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

  = 0.0229 moles + 0.0406 moles

n  = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K  /  0.536 L

P = 1.554 / 0.536

<u><em>P = 2.8999 atm</em></u>

Total pressure in the flask is  2.8999 atm, while assuming the temperature constant.

7 0
2 years ago
Read 2 more answers
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