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2 years ago

PLEASE SOEMBODY HELP ME AND PLEASE HURRY

Chemistry

1 answer:

madreJ [45]2 years ago

7 0

Answer:

I am very sorry I don't have the answer, I just really need the points for more badges.

Explanation:

Also, I really need this answer as well, so can someone please answer, and if someone does answer, please make sure it is at least 500 words.

✨ I hope everyone who sees this has a great day though! ✨

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Use the information in the table to describe the temperature-vs.-time diagrams.

lara31 [8.8K]

Answer:

At 30 and 2,204

diagonal

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top horizontal line

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the change from a solid to a liquid

Explanation:

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3 years ago

What does the exosphere do??

vovikov84 [41]

In the case of bodies with substantial atmospheres, such as Earth's atmosphere, the exosphere is the uppermost layer, where the atmosphere thins out and merges with interplanetary space. It is located directly above the thermosphere.

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4 years ago

Read 2 more answers

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

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 fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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3 years ago

What is the name of Cr3P

almond37 [142]

Answer:

Chromium (I) phosphide.

Explanation:

Hope this help:D

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3 years ago

Please help with the last 2.

ASHA 777 [7]

Answer: Yttruim: 39 protons, 50 neutrons, 39 electrons

Ruthenium: 44 protons, 57 neutrons, 44 electrons

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3 years ago