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Sholpan [36]
3 years ago
9

PPPPPPPLLLLLLLLZZZZZZ HELP I'LL GIVE BRAINLIEST

Chemistry
2 answers:
victus00 [196]3 years ago
8 0

The answer is going to be the Food Web!

Alex17521 [72]3 years ago
7 0

Food web should be the correct answer

:)


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When the following equation is balanced, what is the coefficient of oxygen?
Gwar [14]

Answer:

The answer is 3

C2H5OH + O2 CO2 +H2O (unbalanced)

C2H5OH +3O2(g). 2CO2(g)+3H2O(balanced)

4 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
A chemical reaction produced 10.1 cm3 of nitrogen gas at 23 °C and 746 mmHg. What is the volume of this gas if the temperature a
adell [148]

Answer:

volume of gas = 9.1436cm³

Explanation:

We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.

Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.

V1 =10.1cm³ , P1 =746mmHg,     T1=23°C =23+273=296k

V2 =? ,   P2 =760mmmHg ,     T2=0°C = 0+273 =273K

Using the combined gas equation to calculate for V2;

\frac{V1P1}{T1}=\frac{V2P2}{T2} \\ re-arranging, \\V2 =\frac{V1P1T2}{P2T1}

V2 =\frac{10.1*746*273}{760*296}

V2=9.1436cm³

6 0
3 years ago
Convert 1.71 × 1024 atoms of carbon to moles of carbon.
beks73 [17]
You need to use Avogadro's constant to convert from atoms of carbon to moles of carbon.

1.71*10^24 atoms C * (1 mole/6.022*10^23 atoms) = 2.84 moles of carbon
7 0
3 years ago
What it the oxidation number for cl?<br> A. -2<br> B. +2<br> C. -1<br> D. +1
brilliants [131]

Answer:

c.

Explanation:

it is c.-1 because oxidation number of cl is-1

5 0
3 years ago
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