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elixir [45]
3 years ago
5

How many sig figs are in 0.00205kg

Chemistry
2 answers:
gavmur [86]3 years ago
6 0

Answer:

Three

Explanation:

Significant figures:

The given measurement have three significant figures 2,0,5.

All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

Zero between the non zero digits are consider significant like 104 consist of three significant figures.

The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

fiasKO [112]3 years ago
4 0
It’s 20500000, here are some extras if u need

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A. is <em>incorrect</em>. Scientific laws are so well established that there can be no exceptions.

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Which of the following statements are true about whether atoms tend to gain<br> or lose electrons?
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Answer:

C and D

Explanation:

Atoms with five, six or seven valance electrons gain electrons to complete the octet because it is more convenient for the atoms to gain three, two or one electron as compared to lose five, six or seven electrons. Thus atoms with five, six or seven valance electrons form negative ions by gaining electrons.

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5 0
3 years ago
What do the elements sulfur (s), nitrogen (n), phosphorus (p), and bromine (br) have in common? refer to the periodic table for
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What volume of a 2.5 M stock solution of acetic acid (HC2H3O2) is required to prepare
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<span> </span>

Answer is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
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V</span>₁(CH₃COOH) = ?<span>
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8 0
3 years ago
An unknown compound is processed using elemental analysis and found to contain 117.4g of platinum 28.91 carbon and 33.71g nitrog
dlinn [17]

Answer:

1 mole of platinum

Explanation:

To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.

The empirical formula for the compound can be obtained as follow:

Platinum (Pt) = 117.4 g

Carbon (C) = 28.91 g

Nitrogen (N) = 33.71 g

Divide by their molar mass

Pt = 117.4 / 195 = 0.602

C = 28.91 / 12 = 2.409

N = 33.71 / 14 = 2.408

Divide by the smallest

Pt = 0.602 / 0.602 = 1

C = 2.409 / 0.602 = 4

N = 2.408 / 0.602 = 4

The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.

8 0
3 years ago
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