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Alina [70]
3 years ago
10

PLZ HELP I WILL GIVE BRAINLISTS TO RIGHT ANSWER

Chemistry
1 answer:
galina1969 [7]3 years ago
6 0
..........The answer is B
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When 100ml of M HCl is mixed with 100ml of N NaOH, the pH of the resulting solution is
RoseWind [281]

Answer:

Option d. 7

Explanation:

A mixture of a strong base and a strong acid produce a neutral salt and water.

This is the reaction of neutralization:

HCl + NaOH → NaCl + H₂O

NaCl →  Na⁺  +  Cl⁻

Sodium chloride is neutral salt which does not give H⁻ neither OH⁻ to medium, that's why pH is neutral.

Both ions are derivated from a strong acid and base so they do not make hydrolisis. They are a conjugate pair of a weak acid and base. The reactions can not occur:

Cl⁻  +  H₂O ← OH⁻  +  HCl

Na⁺  +  H₃O⁺ ← NaOH  + H₂O

4 0
3 years ago
Alice and Marge are studying the properties of matter. The girls cut some silver-colored magnesium sheets into 3-inch long strip
frosja888 [35]
<span>B) Magnesium reacts in acids</span>
4 0
3 years ago
Read 2 more answers
A hypothetical covalent molecule, x–y, has a dipole moment of 1.44 d and a bond length of 163 pm. calculate the partial charge o
Aneli [31]
As we know,
                                     1 D  =  3.34 × 10⁻³⁰ C.m
So,
                                     1.44 D  =  ?
Solving for 1.44 D,
                                     =  (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
                    
                         1.44 D  =  4.80 × 10⁻³⁰ C.m

Dipole Moment 
is given as,
 
                         Dipole Moment  =  q  ×  r    
Solving for q,
                         q  =  Dipole Moment / r    ------ (1)
Where,
                         Dipole Moment  =  4.80 × 10⁻³⁰ C.m

                         r  =  163 pm  =  1.63 × 10⁻¹⁰ m

Putting values in eq. 1,

                            q  =  4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m

                            q  =  2.94 × 10⁻²⁰ C

As,
                            1.602 × 10⁻¹⁹ C  =  1 e⁻
So,
                             2.94 × 10⁻²⁰ C  =  X e⁻

Solving for X,

                            X  =  (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C

                                = 0.183 e⁻

Result:
           
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
4 0
3 years ago
An automobile tire contains air at 320.×103 Pa at 20.0 ◦C. The stem valve is removed and the air is allowed to expand adiabatica
NISA [10]

Answer:

6.15.3 k

Explanation:

From the question we can see that

q = 0,  Δu = w

Then,

T_f = \frac{C_{V,m}+RP_{ext}P_i}{C_{V,m}+RP_{ext}P_f} T_i

putting values wet

=\frac{2.5\times 8.314+8.314\left(10^5\right)\left(3.20\times 10^5\right)}{2.5\times 8.314+\left(8.314\right)\left(10^5\right)\left(10^5\right)}\times \:293

T_f = 615.3 K

6 0
3 years ago
Determine the pH of a 5x10^-4 M solution of Ca(OH)2
miss Akunina [59]
Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-   

Ca(OH)</span>₂ is <span>strong Bases</span><span>

</span>Therefore,  the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]

pOH = - log [ <span>5 x 10⁻⁴ ]

pOH = 3.30

pH + pOH = 14

pH + 3.30 = 14

pH = 14 - 3.30

pH = 10.7

hope this helps!</span>
5 0
3 years ago
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