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Goryan [66]
3 years ago
7

What kind of chemical reaction is below? AgNO3 + NaCI -> AgCI + NaNO3

Chemistry
2 answers:
andrezito [222]3 years ago
7 0
Reaction is the Double Replacement or Displacement reaction, depending on what textbook you use in school. :) 

AX+ BY -------> AY+BX

i'm going to insert a picture to show you

frozen [14]3 years ago
6 0
A double replacement reaction is happening in that equation.

Here is the formula for double replacement:

AB+CD ---> AD+CB
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2 years ago
How many liters are equivalent to 645 mililiters
abruzzese [7]

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0.645 liters

Explanation:

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6 0
4 years ago
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What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
blsea [12.9K]
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value  = 5.3 x 10^-4

And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

                    =0.00516 M 

Thus, pOH = 2.30 
 pH = 14 - pOH = 11.7
6 0
3 years ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
slava [35]

Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation  

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA  = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

a = 9.673 * 10^-7

OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

pH = 14-6.02 = 7.98

8 0
3 years ago
A 0.175 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.27° in a 1-dm sample
-BARSIC- [3]

Answer:

The specific rotation of D is 11.60° mL/g dm

Explanation:

Given that:

The path length (l) =  1 dm

Observed rotation (∝) = + 0.27°

Molarity = 0.175 M

Molar mass = 133.0 g/mol

Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol

Concentration in (g/mL) = 23.275 g/L

Since 1 L = 1000 mL

Concentration in (g/mL) = 0.023275 g/mL

The specific rotation [∝] = ∝/(1×c)

= 0.27°/( 1  dm ×  0.023275 g/mL )

= 11.60° mL/g dm

Thus, the specific rotation of D is 11.60° mL/g dm

3 0
3 years ago
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