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3 years ago

Poh of a solution with a ph of 5.00

Chemistry

1 answer:

creativ13 [48]3 years ago

8 0

Answer:

Explanation:

14-5.=pOH

- Hope that helps! Please let me know if you need further explanation.

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Electron configurations are a shorthand form of an orbital diagram, describing which orbitals are occupied for a given element.

kipiarov [429]

Answer:

[Ar] 3d10 4s2 4p3 is shorthand, 1s22s22p63s23p63d104s24p3 long hang

Explanation:

The shorthand is made using the lowest & closest noble gas, and picking up where it leaves off as follows, and longhand is made from a followed pattern you can easily find

3 0

3 years ago

Read 2 more answers

Find the density of an 8 gram rock if the water in a graduated cylinder rises from 25.0 mL to

Olin [163]

To measure the density of the stone placed in a graduated cylinder let us follow these steps bellow

Measure the volume of water poured into a graduated cylinder
Place the object in the water and remeasure the volume.
The difference between the two volume measurements is the volume of the object.
Divide the mass by the volume to calculate the density of the object.

We know that the formula for density is given as

Given data

Mass = 8gram

Initial Volume of water in cylinder = 25mL

Final Volume of water in cylinder = 29mL

Hence the volume of the rock = 29-25 = 4mL

Therefore the density of the rock = 8/4 = 2 g/mL

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6 0

2 years ago

Mountains and valleys are some of the features found in the layer of rock at Earth's surface called the

loris [4]

Answer:

Narrow

Explanation:

Narrow

5 0

3 years ago

Calculate the number of moles of carbon dioxide formed when 40.0 moles of oxygen is consumed in the burning of propane. Note the

joja [24]

The first step is to balance the equation:

C3H8 + 5O2 ---> 3CO2 + 4H2O

Check the balance

element left side right side

C 3 3
H 8 4*2 = 8
O 5*2=10 3*2 + 4 = 10

Then you have the molar ratios:

3 mol C3H8 : 5 mol O2 : 3 mol CO2 : 4 mol H2O

Now you have 40 moles of O2 so you make the proportion:

40.0 mol O2 * [3 mol CO2 / 5 mol O2] = 24.0 mol CO2.

Answer: option D. 24.0 mol CO2

7 0

3 years ago

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

6 0

3 years ago