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3 years ago

PLEASE HELP FAST!!!!

Chemistry

1 answer:

deff fn [24]3 years ago

6 0

Answer:

2c +o2=2co

Explanation:

Check and u will see that all the elements in the left side of the chemical reaction are equal to those on the right

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natima [27]

It should be C but not sure

7 0

3 years ago

NEED HELP! FAST! WILL CHOOSE BRAINLIEST ANSWER!

Jobisdone [24]

Mechanical waves need mediums. or else it isn't a mechanical wave

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]

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.

8 0

3 years ago

What is the mole fraction of oxygen in a gaseous mixture containing 25 grams of oxygen, 15 grams of nitrogen, and 10 grams of he

4vir4ik [10]

Convert all the gases from grams to moles using their molar masses first. Remember that nitrogen and oxygen exist as DIATOMIC gases.

O2 - 25 g / 32 g/mol = 0.78 mol O2
N2 - 15 g / 28.02 g/mol = 0.54 mol N2
He - 10 g / 4.0 g/mol = 2.5 mol He

Add all the moles of gases.

0.78 + 0.54 + 2.5 = 3.82 moles

Divide O2 moles by total moles.

0.78/3.82 = 0.20

7 0

3 years ago

Given the balanced ionic equation:

bezimeni [28]

Answer : The correct option is, (4) 6.0 mol

Explanation :

The given balanced chemical equation is,

$3Pb^{2+}+2Cr\rightarrow 3Pb+2Cr^{3+}$