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Marina86 [1]
4 years ago
8

ILL GIVE YOU BRAINLIST * have to get it right * !!! Although individual body systems can perform specific functions, they depend

on one another and work together for the good of the entire organism.
Which of the following provides the best evidence that the digestive system and the muscular system work together to carry out their functions?

Chemistry
1 answer:
andre [41]4 years ago
7 0
I think it might be A muscles help move nutrients through the body to the stomach
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Please help!!! Best answer will get brainliest
Effectus [21]
The answer should be metallic
5 0
3 years ago
Read 2 more answers
A student wishes to prepare 300-mL of a 0.180 M potassium bromide solution using solid potassium bromide, a 300-mL volumetric fl
givi [52]

Answer:

a) 6.426  grams of potassium bromide student must weigh.

b) The correct answer is option a.

Explanation:

Concentration(C)=\frac{Moles(n)}{Volume (L)}

a) Concentration of potassium bromide = c = 0.180 M

Moles of potassium bromide = n

Volume of the  potassium bromide solution = 300 mL = 0.3 L

0.180 M=\frac{n}{0.3 L}

n=0.180 M\times 0.3 L=0.054 mol

Mass of 0.054 moles of potassium bromide :

= 0.054 mol\times 119 g/mol=6.426 g

6.426  grams of potassium bromide student must weigh.

b) Drying of flask in drying oven. This is because solution will be prepared by adding water after adding solute. So drying the flask will be not required.

5 0
4 years ago
Read 2 more answers
What happens to items that have a lower density than water are submerged in water
sergiy2304 [10]

Those items float.

Because the items are less dense than water, they float. If they are more dense than water, they sink.

5 0
3 years ago
Read 2 more answers
Calculate how many grams of iron can be made from 16.5 grams of fe2o3 and 10 grams of h2 by fe2o2 3h2 to 2fe 3h2o find the limit
Lisa [10]
How can I help with this question?

7 0
3 years ago
Calculate the percent ionization of a 0.18 M benzoic acid solution in a solution containing 0.10 M sodium benzoate.
Yuliya22 [10]

Answer:

% I = 0.083 %

Explanation:

  • C6H5COOH  +  NaOH ↔ NaC6H5CO2 + H2O

∴ M C6H5COOH = 0.18 mol/L

∴ M NaC6H5CO2 = 0.10 mo/L

  • % ion = ( { H3O+ ] / initial acid concentration ) * 100

⇒ C6H5COOH ↔ H3O+  +  C6H5COO-

⇒ NaC6H5CO2 ↔ Na+  +  C6H5COO-

∴ Ka = 6.4 E-5 = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ]  

Ka value taken from the literature

mass balance

⇒ 0.18 + 0.1 = [ C6H5COOH ] + [ C6H5COO- ]

⇒ [ C6H5COOH ] = 0.28 - [ C6H5COO- ]  ........(1)

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ C6H5COO- ]

∴ [ Na+ ] ≅ M NaC6H5CO2 = 0.1 M

⇒ [ C6H5COO- ] = [ H3O+ ] + 0.1..............(2)

(1) and (2) in Ka:

⇒ 6.4 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 0.1 ) / ( 0.28 - ( [ H3O+ ] + 0.1 ) )

⇒ 6.4 E-5 = [ H3O+ ]² + 0.1 [H3O+ ] / ( 0.18 - [ H3O+ ] )

⇒ 1.17 E-5 - 6.5 E-5 [ H3O+ ] = [ H3O+ ]² + 0.1 [ H3O+ ]

⇒ [ H3O+ ]² + 0.1 [ H3O+ ] - 1.17 E-5 = 0

⇒ [ H3O+ ] = 1.493 E-4 M

⇒ % I = ( 1.493 E-4 / 0.18 ) * 100 = 0.083 %

6 0
3 years ago
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