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attashe74 [19]
3 years ago
7

A 10 ft diameter by 15 ft height vertical tank is receiving water (p 62.1 lb/ cu ft) at the rate of 300 gpm and is discharging t

hrough a 6 in ID line with a constant speed of 5 fps. At a given instant, the tank is half full. Find the water level and the mass change in the tank 15 min later.
Chemistry
1 answer:
Alchen [17]3 years ago
8 0

Answer:

werrrrrrrrrrrrrrr

Explanation:

You might be interested in
17 moles of oxygen is equal to
mixas84 [53]

Answer:

271.9897999999996

Explanation:

have a good day :)

3 0
3 years ago
Natural substances that have characteristic properties and somewhat specific chemical compositions are called ______.
Darya [45]

Answer:

Minerals

Explanation:

Minerals are usually defined as naturally occurring inorganic substances, that are comprised of specific chemical composition as well as having a well organized internal structure of atoms.

Different mineral forms under different conditions and under different geological settings. These minerals combine with one another giving rise to the formation of rocks.

They are characterized by the presence of both physical as well as chemical properties.

Example of some minerals includes, Calcite, Fluorite, Quartz, and Feldspar.

8 0
3 years ago
What happens in a single-replacement reaction?
antoniya [11.8K]

Answer:

D

Explanation:

A single-replacement reaction occurs when one element replace another in a single compound.

8 0
3 years ago
Chromium has an atomic mass of 51.9961 u and consists of four isotopes, 50 Cr , 52 Cr , 53 Cr , and 54 Cr . The 52 Cr isotope ha
Troyanec [42]

Answer : The atomic mass of _{24}^{53}\textrm{Cr} isotope is 52.8367 amu

Explanation :

We know that:

Total percentage abundance of the isotope = 100 %

Percentage abundance of _{24}^{50}\text{Cr}\text{ and }_{24}^{53}\textrm{Cr}\text{ isotopes}=[100-(83.79+2.37)]=13.84\%

We are given:

Ratio of _{24}^{50}\textrm{Cr}\text{ and }_{24}^{53}\textrm{Cr} isotopes = 0.4579 : 1

Percentage abundance of _{24}^{50}\textrm{Cr} isotope = \frac{0.4579}{(0.4579+1)}\times 13.84\%=4.37\%

Percentage abundance of _{24}^{53}\textrm{Cr} isotope = \frac{1}{(0.4579+1)}\times 13.84\%=9.49\%

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 mass of _{24}^{53}\textrm{Cr} isotope be 'x'

For _{24}^{50}\textrm{Cr} isotope:

Mass of _{24}^{50}\textrm{Cr} isotope = 49.9460 amu

Percentage abundance of _{24}^{50}\textrm{Cr} = 4.37 %

Fractional abundance of _{24}^{50}\textrm{Cr} isotope = 0.0437

For _{24}^{52}\textrm{Cr} isotope:

Mass of _{24}^{52}\textrm{Cr} isotope = 51.9405 amu

Percentage abundance of _{24}^{52}\textrm{Cr} isotope = 83.79 %

Fractional abundance of _{24}^{52}\textrm{Cr} isotope = 0.8379

or _{24}^{53}\textrm{Cr} isotope:

Mass of _{24}^{53}\textrm{Cr} isotope = x amu

Percentage abundance of _{24}^{53}\textrm{Cr} isotope = 9.49 %

Fractional abundance of _{24}^{53}\textrm{Cr} isotope = 0.0949

For _{24}^{54}\textrm{Cr} isotope:

Mass of _{24}^{54}\textrm{Cr} isotope = 53.9389 amu

Percentage abundance of _{24}^{54}\textrm{Cr} isotope = 2.37 %

Fractional abundance of _{24}^{54}\textrm{Cr} isotope = 0.0237

Average atomic mass of chromium = 51.9961 amu

Putting values in equation 1, we get:

51.9961=[(49.9460\times 0.0437)+(51.9405\times 0.8379)+(x\times 0.0949)+(53.9389\times 0.0237)]\\\\x=52.8367amu

Hence, the atomic mass of _{24}^{53}\textrm{Cr} isotope is 52.8367 amu

4 0
4 years ago
Given the net ionic equation, Ba2+ + SO42- => BaSO4, how many grams of barium chloride must be present to react with 200 gram
zheka24 [161]

Answer: 208g/mol

Explanation:

First of all we have to write the balance equation for the reaction.

BaCl2 + Fe2(SO4)3____> BaSO4 + FeCl3

After balancing we have.

3BaCl2 +Fe2(SO4)3_____> 3BaSO4 +2FeCl3

Looking at the equation, we find out that 3 moles of barium chloride reacts with 1 mole of iron iii sulfate

Therefore we have

3moles of BaCl2 _____> 400g/mole of iron iii sulfate

Xmole of BaCl2 _____> 200g/mole of iron iii sulfate

X = 2 * 200g/mol divide by 400g/mol

X = 1mole

1 mole of BaCl2 will be need to react with 200g/mol of iron iii sulfate.

This 1 mole of BaCl2 is equivalent to 208g/mol of BaCl2.

Therefore the gram of barium chloride that must be present is = 208g/mol//

3 0
4 years ago
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