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Nookie1986 [14]
3 years ago
10

state the rules governing the number of significant figures that result from each of the following operations addition and subtr

action; mutiplication and divison
Chemistry
1 answer:
lubasha [3.4K]3 years ago
4 0
Adding and subtracting it's how many are behind the decimal, and multiply its how many digits
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Solid magnesium has mass of 1300g and a volume of 743cm . what's the density of magnesium?
Lilit [14]
D = m / V

d = 1300 g / 743 cm³

d = 1.749 g/cm³
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2 years ago
Question 4 of 15
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B.

Explanation:

5 0
2 years ago
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If a recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb what volume in liters
Bad White [126]

The volume in liters of 3245 aluminum cans is 24.8 L.

A recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb. The mass of 3245 aluminum cans is:

3245 cans \times \frac{1lb}{22cans} = 147.5 lb

To convert mass to volume, we need the density of aluminum (5.95 lb/L). The volume corresponding to 147.5 lb of aluminum is:

147.5 lb \times \frac{1L}{5.95 lb} = 24.8 L

The volume in liters of 3245 aluminum cans is 24.8 L.

You  can learn more about density here: brainly.com/question/1841285

3 0
2 years ago
What is the current model of atom
leonid [27]

Answer:

The bohr model is the model in use today

7 0
2 years ago
In a typical analysis, 15 ml of an aqueous solution containing an unknown amount of acetylcholine had a ph of 7.65. When incubat
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pH of the acetyl choline solution before incubation = 7.65

[H_{3}O^{+}]=10^{-7.65}=2.24*10^{-8}M

pH of the solution after incubation = 6.87

[H_{3}O^{+}]=10^{-6.87}=1.35*10^{-7}M

The difference in concentration of hydronium ion before and after incubation

=1.35*10^{-7}M-2.24*10^{-8}M=1.126*10^{-7}M

This difference in hydronium ion concentration can be attributed to the increase in the concentration of acetic acid, which is formed when acetylcholine is hydrolyzed by acetycholinesterase. The mole ratio of acetylcholine to acetic acid is 1:1.

Therefore the moles of acetylcholine = 15 mL * \frac{1L}{1000mL}*\frac{1.126*10^{-7}mol }{L}=1.689*10^{-9}mol


7 0
2 years ago
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