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GREYUIT [131]
3 years ago
12

Which species is most vulnerable to extinction raccoon or mountain gorilla

Chemistry
2 answers:
777dan777 [17]3 years ago
6 0
Mountain Gorilla’s are more vulnerable to extinction because they are endangered due to habitat loss, diseases, poaching, ect.
sineoko [7]3 years ago
4 0
Mountain gorilla because they are endangered
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Mercury, which is sometimes used in thermometers, has a density of 13.534 g/ml at room temperature. what volume of mercury conta
Sergeu [11.5K]
D - density: 13,534 g/ml
m - mass: 10g
V - volume: ??
_____________
d = m/V
V = m/d
V = 10/13,534
V = 0,7389 ml

:•)
6 0
3 years ago
Which option correctly describes the influence of electronegativity difference on bond character?
Bezzdna [24]
Electronegativity is a property of a certain element to attract an electron from other elements towards itself during a reaction. The greater the difference of the electronegativities of two elements in a compound, the stronger is their binding force. The strongest bond is the ionic bond, then following by covalent bonds.
7 0
3 years ago
What is science and why we use science ​
oee [108]

Answer:

YES

Explanation:

WE use science cause it is all aaprindhs and science us everywhere .

5 0
3 years ago
Read 2 more answers
Question 9
Shkiper50 [21]

Answer: There are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

Explanation:

Given: M_{1} = 1.0 M,    V_{1} = ?

M_{2} = 0.1 M,    V_{2} = 200 mL

Formula used is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute values into the above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\1.0 M \times V_{1} = 0.1 M \times 200 mL\\V_{1} = \frac{0.1 M \times 200 mL}{1.0 M}\\= 20 mL

Thus, we can conclude that there are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

7 0
3 years ago
a certain anesthetic contains 64.9% C, 13.5% H, and 21.6% O by mass. at 120 deg Celsius & 750 mmHg, 1.00 L of the gaseous co
slamgirl [31]
You need to use the % information to determine the empirical formula of the compound first. 

The empirical formula is the simplest ratio of atoms in the molecule. 

Then use the rest of the data to determine moles of gas, and use this to determine molar mass of gas... 

Empirical formula calculations:

Assume you have 100 g, calculate the moles of each atom in the 100 g 

moles = mass / molar mass 
molar mass C = 12.01 g/mol 
molar mass H = 1.008 g/mol 
molar mass O = 16.00 g/mol 

C = 64.9 % = 64.6 g 
H = 13.5 % = 13.5 g 
O = 21.6 % = 21.6 g 

moles C = 64.6 g / 12.01 g/mol = 5.38 mol 
moles H = 13.5 g / 1.008 g/mol = 13.39 mol 
moles O = 21.6 g / 16.00 g/mol = 1.35 mol 

So ratio of C : H : O 
is 5.38 mol : 13.39 mol : 1.35 mol 

Divide each number in the ratio by the lowest number to get the simplest whole number ratio 

(5.38 / 1.35) : (13.39 / 1.35) : (1.35 / 1.35) 

4 : 10 : 1 

empirical formula is 
C4H10O 


Finding moles and molar mass calcs 

Now, you know that at 120 deg C and 750 mmHg that 1.00L compound weighs 2.30 g. 

We can use this information to determine the molar mass of the gas after first working out how many moles the are in the 1.00 L 

PV = nRT 
P = pressure = 750 mmHg 
V = volume = 1.00 L 
n = moles (unknown) 
T = temp in Kelvin (120 deg C = (273.15 + 120) Kelvin) 
- T = 393.15 Kelvin 
R = gas constant, which is 62.363 mmHg L K^-1 mol^-1 (when your P is in mmHg and volume is in L) 

n = PV / RT 
n = (750 mmHg x 1.00 L) / (62.363mmHg L K^-1 mol^-1 x 393.15 K) 
n = 0.03059 moles of gas 

We know moles = 0.03509 and mass = 2.30 g 
So we can work out molar mass of the gas 

moles = mass / molar mass 
Therefore molar mass = mass / moles 
molar mass = 2.30 g / 0.03059 mol 
= 75.19 g/mol 


Determine molecular formula 

So empirical formula is C4H10O 
molar mass = 75.19 g/mol 

To find the molecular formula you divide the molar mass by the formula weight of the empirical formula... 
This tells you how many times the empirical formula fits into the molecular formula. Tou then multiply every atom in the empirical formula by this number 

formula weight C4H10O = 74.12 g/mol 

Divide molar mass by formula weight empirical 
75.15 g/mol / 74.12 g/mol 
= 1 
(It doesn't matter that the number don't quite match, they rarely do in this type of calc (although I could have made a slight error somewhere) but the numbers are very close, so we can say 1.) 

The empirical formula only fits into the molar mass once, 

molecular formula thus = empirical formula 
<span>
C4H10O

Therefore, the </span>molecular formula of the compound is <span>C4H10O.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!</span>
5 0
3 years ago
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