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Oxana [17]
3 years ago
10

PLZ ANSWER NOW

Chemistry
2 answers:
abruzzese [7]3 years ago
8 0
The answer is A, hope that helped :)
harina [27]3 years ago
7 0

Answer:

The answer is A.

Explanation:

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The quantity, 1,385 mg is equivalent to in scientific notation
Agata [3.3K]

Answer:

The answer is 1385 x10∧ -3

Explanation:

Scientific notation is a quick way to represent a number using notation in exponential form or powers of base ten.

This notation allows us to express too large or small numbers easily.

<u>For example: </u>

500 is written as 5x10∧2; Where the power 2 represents the number of 0's that follow 5.

0.0093 is written as 9.3x10∧-3; Where the power -3 represents the number of times the comma was moved to the right.

8 0
3 years ago
Help me its due soon thanks
UkoKoshka [18]

Answer:

B

Explanation:

The gravity accelleration values are the same for both planets.

5 0
4 years ago
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How many doubtful digit(s) is/are allowed in any measured quantity?
Mnenie [13.5K]

Answer:

zero

Explanation:

I I think one should be so accurate with measurements and experiments

3 0
3 years ago
Two isotopes of hypothetical element X exist with abundances of 30.00% 100X and 70.00% 101X. What is the approximate atomic mass
Rasek [7]

Answer:

C. 100.7 amu

Explanation:

Isotopes of an element are atoms of an element with the same atomic number but different atomic masses. Each atomic mass of an isotope is known as an isotopic mass. An element that exhibits isotope, that is, that have two or more isotopes has a relative atomic mass that is not a whole number.

Relative atomic mass of X is the sum of the products of the relative abundances of each isotope and its isotopic mass.

For Isotope ¹⁰⁰X: 30% × 100 = 30 amu

For Isotope ¹⁰¹X: 70% × 101 = 70.7 amu

Relative atomic mass of X = (30 + 70.7) amu = 100.7 amu

Therefore, the approximate atomic mass of X is 100.7 amu

4 0
3 years ago
Chlorine is the most widely used disinfectant for killing pathogens during water treatment. Determine the kilograms of chlorine
makkiz [27]

Explanation:

The given data is as follows.

    Flow of chlorine (Q) = 10,000 m^{3}/day

Amount of liter present per day is as follows.

                  10,000 \times 10^{3} l/day

It is given that dosage of chlorine will be as follows.

                   10 mg/l = 10 \times 10^{-6} kg/l

Therefore, total chlorine requirement is as follows.

          Total chlorine requirement = (10,000 \times 10^{3}) \times (10 \times 10^{-6}) kg/day

                                        = 100 kg/day

Thus, we can conclude that the kilograms of chlorine used daily at the given water treatment plant is 100 kg/day.

4 0
3 years ago
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