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2 years ago

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Write the net ionic equation for the reaction between hydrocyanic acid and potassium hydroxide. Do not include states such as (a

q) or (s).

1 answer:

eimsori [14]2 years ago

4 0

Answer:

The net ionic equation is as follows:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

Explanation:

The reaction between Hydrocyanic acid, HCN, and sodium hydroxide is a neutralization reaction between a weak acid and a strong base.

Hydrocyanic acid being a weak acid ionizes only slightly, while sodium hydroxide being a strong base ionizes completely. The equation for the reaction is given below:

A. HCN(aq) + NaOH-(aq) ----> NaCN(aq) + H2O(l)

Since Hydrocyanic acid is written in the aqueous form as it ionizes only slightly and the ionic equation is given below:

HCN(aq) + Na+(aq)+OH-(aq) ----> Na+(aq)+CN-(aq) + H2O(l)

Na+ being a spectator ion is removed from the net ionic equation given below:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

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Fill in the blanks using terms from this unit.

kolbaska11 [484]

Answer:

See the answer below

Explanation:

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2 years ago

Read 2 more answers

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

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2 years ago

Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transit

Lapatulllka [165]

<u>Answer:</u> The wavelength of spectral line is 656 nm

<u>Explanation:</u>

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 2

$n_i$ = Initial energy level = 3

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.524\times 10^6m^{-1}}=6.56\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.56\times 10^{-7}m\times (\frac{10^9nm}{1m})=656nm$

Hence, the wavelength of spectral line is 656 nm

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2 years ago

What is the ph of a 0.15 m solution of ammonium chloride?

Thepotemich [5.8K]

Best Answer: <span>(a)
8.9 x 10^-7 = x^2 / 0.15-x
x = [OH-] = 0.00037 M
pOH = 3.4
pH = 14 - 3.4 = 10.6

(b)
Ka = Kw/Kb = 5.6 x 10^-10 = x^2 / 0.20-x
x = [H+] = 0.000011 M
pH = 5.0</span>

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3 years ago

Write three sentences about The Amazing Life of Sand<br> HELP ASAP PLS

pishuonlain [190]

Answer:“If we’ve covered all of the potential sources, and we know the unique signature of the sand from these different sources, and we find it on a beach somewhere, then we basically know where it came from,” explained Barnard.

Explanation:

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2 years ago