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4 years ago

Hey um i dont get this yall wont to help me out???

Chemistry

1 answer:

den301095 [7]4 years ago

4 0

3. 9
20. 60 (1hour)
To get to 60 from 20 we multiply by three. 3 times 3 is 9.
A is true
B is true
C is true
D is true
E is false

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Which commercial technology commonly uses plasmas?

ankoles [38]

Answer:

<h3>A television is commercial technology commonly uses plasmas. </h3>

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3 years ago

What is the atomic mass of X ? Formula of the compound is XO3? Also identify the element X ? The molecular mass of XO3 is 80? (

djyliett [7]

Answer:

Explanation:

XO3

X + 3x16 = 80

x= 32 atomic mass for sulfur

SO3

6 0

3 years ago

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]

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

<u>Explanation:</u>

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'

<u>For $_{77}^{191}\textrm{Ir}$ isotope:</u>

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

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

<u>For $_{77}^{193}\textrm{Ir}$ isotope:</u>

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.

8 0

3 years ago

Help me pretty please

Natalka [10]

Answer:

The top answer

7 0

3 years ago

13.3 g of benzene (C6H6) is dissolved in 282 g of carbon tetrachloride. What is the molal concentration of benzene in this solut

Verizon [17]

Answer:

0.605 molal

Explanation:

molality is the amount of solute in a particular mass of solvent.

lets calculate the amount of benzene solute.

mass of benzene= 13.3g

molar mass of C6H6= 12*6 +1*6 =72+7=78g/mol

amount of benzene= mass/molar mass

=13.3/78

=0.1705mol

molality= amount of solute/mass of solvent in kg

mass of solvent=282g=0.282kg

molality = 0.1705/0.282

=0.605 molal

6 0

3 years ago