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viktelen [127]
3 years ago
5

HELP O DONT UNDERSTAND THIS!IM SOOO STRESSED HOW DO I SOLVE IT

Chemistry
1 answer:
katrin [286]3 years ago
4 0

Answer:

.86

.48

Explanation:

19/25 x 4/4 = 86/100

86/100 = .86

12/25 x 4/4 = 48/100

48/100 = .48

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Mashutka [201]
A subduction zone<span> is a region of the Earth's crust where tectonic plates meet. Tectonic plates are massive pieces of the Earth's crust that interact with each other. The places where these plates meet are called plate boundaries.

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7 0
4 years ago
Read 2 more answers
Element Y has two natural isotopes Y-63 (62.940 amu) and Y-65 (64.928 amu). Calculate the atomic mass of element Y, given the ab
djverab [1.8K]

Answer:

Average atomic mass = 63.553 amu.

Explanation:

Given data:

Abundance of Y-63 = 69.17%

Abundance of Y-65 = 100 - 69.17 = 30.83%

Atomic mass of Y-63 = 62.940 amu

Atomic mass of Y-65 = 64.928 amu

Atomic mass of Y = ?

Solution:

Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass= (62.940×69.17)+(64.928×30.83) /100

Average atomic mass =  4353.560 + 2001.730 / 100

Average atomic mass = 6355.29 / 100

Average atomic mass = 63.553 amu.

5 0
4 years ago
Why were the sample of salt and sand placed into 50-ml beakers prior to weighing?
LUCKY_DIMON [66]

Answer : Salt and Sand are weighed and kept in different 50 Milliliter beakers because, the experiment is about studying and separating the homogeneous mixture. As salt crystals are cubic and sand grains are rounded oblong grains. If we mix it together it would require a physical method to separate them. So, before starting the experiment it needs to be weighed in different beakers.

7 0
3 years ago
What are the physical properties of toothpaste
bogdanovich [222]
The whitening toothpastes showed differences in their physical-chemical properties. All toothpastes promoted changes to the surface, probably by the use of a bleaching agent.
5 0
4 years ago
In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole pro
Ksju [112]

Based on experiment 1:

Mass of Hg = 1.00 g

Mass of sulfide = 1.16 g

Mass of sulfur = 1.16 - 1.00 = 0.16 g

# moles of Hg = 1 g/200 gmol-1 = 0.005 moles

# moles of S = 0.16/32 gmol-1 = 0.005 moles

The Hg:S ratio is 1:1, hence the sulfide is HgS

Based on experiment 2:

Mass of Hg taken = 1.56 g

# moles of Hg = 1.56/200 = 0.0078

Mass of S taken = 1.02 g

# moles of S = 1.02/32 = 0.0319

Hence the limiting reagent is Hg

# moles of Hg reacted = # moles of HgS formed = 0.0078 moles

Molar mass of HgS = 232 g/mol

Therefore, mass of HgS formed = 0.0078 * 232 = 1.809 g = 1.81 g

3 0
3 years ago
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