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Pachacha [2.7K]
2 years ago
11

i have a crossword thing that i got from school and one of the clues is: If there are two equal and opposite forces on something

, we say the forces are______ .Its 8 letters and the 4th letter is A. What do you think it is im so confused?
Chemistry
2 answers:
Nezavi [6.7K]2 years ago
8 0
Well the forces are balanced. Perhaps the 8th letter is not actually "A" but a "D"? 
cluponka [151]2 years ago
6 0
I would assume the answer is called attractive
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When Aluminum foil is added to a solution of Copper (II) chloride (blue color solution), the following reaction occurs.
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The volume of a pond being studied for the effects of acid rain is 35 kiloliters (kL). There are 1,000 liters (L) in 1 kL and 1
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35,000,000,000 mL

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3 years ago
Which bond is most polar c-o h-o n-o s-o
scZoUnD [109]

Answer:

Option 4 with o-h in the most polar bond, since the two atoms in the bond have the greatest difference in electronegativity.  This is assuming there are no other factors in other atoms bound to either of the elements in the bond.

Explanation:

6 0
3 years ago
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
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