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marusya05 [52]
3 years ago
14

It pulls everything down towards earth

Chemistry
1 answer:
irga5000 [103]3 years ago
5 0
The answer is gravity:
Explanation: an invisible force that pulls objects toward each other. Earth's gravity is what keeps you on the ground and what makes things fall. Anything that has mass also has gravity.
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How to unscramble dtneinreigs
torisob [31]
ANSWER: Ingredients.
3 0
3 years ago
Read 2 more answers
How do many types of cells, like bacteria, reproduce?
umka2103 [35]

Answer:

cells like bacteria are bisexual so they split and their offspring is 100 percent like the parent and this process happens over and over

Explanation:

8 0
2 years ago
50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the
scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

H_2NC_2H_5CO^-_2  +  H^+  ------>  H_3}^+NC_2H_5CO^-_2

1 mole of  alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL  of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via  the following process as shown below;

[H_3}^+NC_2H_5CO^-_2] = \frac{initial moles of alaninate}{total volume}

[H_3}^+NC_2H_5CO^-_2] = \frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}

[H_3}^+NC_2H_5CO^-_2] = \frac{8}{100mL}

[H_3}^+NC_2H_5CO^-_2] = 0.08 M

Alanine serves as an intermediary form, however the concentration of H^+ and the pH can be determined as follows;

[H^+] = \sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{  K_{a1}{[H_3}^+NC_2H_5CO^-_2]  } }

[H^+] = \sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})}  {(10^{-pK_{a1}})+(0.08)} }

[H^+] = \sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})}  {(10^{-2.344})+(0.08)} }

[H^+] =  7.63*10^{-7}M

pH = - log [H^+]

pH = -log[7.63*10^{-7}]

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

H_2NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO^-_2

H^+_3NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO_2H

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

[H^+_3NC_2H_5CO_2H] = \frac{initial moles of alaninate}{total volume}

[H^+_3NC_2H_5CO_2H] = \frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}

[H^+_3NC_2H_5CO_2H] = \frac{8}{150}

[H^+_3NC_2H_5CO_2H] = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

H^+_3NC_2H_5CO_2H        ⇄        H^+_3NC_2H_5CO^-_2    +  H^+

(0.053 - x)                                  x                             x

K_{a1} = \frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}

10^{-PK_{a1}} = \frac{x*x}{(0.053-x)}

10^{-2.344} =\frac{x^2}{(0.053-x)}

0.00453 = \frac{x^2}{(0.053-x)}

0.00453(0.053-x) =x^2

x^2+0.00453x-(2.4009*10^{-4})

Using quadratic equation formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

we have:

\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)} OR \frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}

= 0.0134                    OR                -0.0179

So; we go by the positive integer which says

x = 0.0134

So [H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M

pH = -log[H^+]

pH = -log[0.0134]

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0
3 years ago
How can we calculate empirical formula​
Serggg [28]

Explanation:

TO calculate emperical formula , we need percentage composition of the elements and molecular mass of a specific compound which is already given in the question

then create a table with 1.name of element, 2.percentage given and its atomic mass , 3.no. of moles

( %given/at.mass of element) ... then take the least value in the no. of moles you have found now and divide with each no. of moles value .. to get a simple digit number.

You have got the empirical formula

7 0
3 years ago
For 2Na (s) + Cl2(g) - 2NaCI (s) ,how many grams of sodium Chloride will be made from 115 g of sodium metal when reacted with ex
Solnce55 [7]

<u>Given:</u>

Mass of Na = 115 g

Excess Cl2

<u>To determine:</u>

Mass of NaCl produced

<u>Explanation:</u>

Given reaction is-

2Na(s) + Cl2(g) → 2NaCl(s)

Since Cl2 is in excess, Na will be the limiting reagent

As per the reaction stoichiometry Na:NaCl = 1:1

i.e. moles of Na reacted = moles of NaCl formed

Now, # moles of Na = mass of Na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

Therefore, moles of NaCl = 5

Molar mass of NaCl = 58 g/mol

Mass of NaCl = 5 moles * 58 g.mol-1 = 290 g

Ans: Amount of Nacl produced = 290 g

5 0
3 years ago
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