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alukav5142 [94]
4 years ago
5

Iron nails do not get rusted when kept in distilled water even for a long time. Give reason

Chemistry
1 answer:
barxatty [35]4 years ago
5 0
Rust is oxidation of the iron surface.  It takes oxygen too,
of which there is none available in the jar of water.
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Emission Spectrum Questions for Quiz Print - Quizizz
-Dominant- [34]

Isotopes of  the same element

₃₅⁷⁷X and ₃₅⁸¹X

<h3>Further explanation</h3>

Given

Isotopes of element

Required

Isotopes of  the same element

Solution

The elements in nature have several types of isotopes  

Isotopes are elements that have the same Atomic Number (Proton)  and different mass numbers

Element symbols that meet the 2 conditions above are :

₃₅⁷⁷X and ₃₅⁸¹X

3 0
3 years ago
If a sample contains 21.2g N how many moles of N does it contain
tigry1 [53]

Hey there!

The molar mass of nitrogen is 14.007.

Convert grams to moles:

21.2 ÷ 14.007 = 1.51

The sample containing 21.2g of nitrogen contains 1.51 moles of nitrogen.

Hope this helps!

8 0
3 years ago
Ductile means...
balu736 [363]

Answer:

A

Explanation:

3 0
3 years ago
PH is 7.45. Calculate value of [H3O+] and [OH-]
lana66690 [7]

Answer:

The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M

Explanation:

We use the formulas:

pH= - log(H30+)  and Kwater=(H30+)x(OH-)

pH= - log(H30+)  ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M

Kwater=(H30+)x(OH-)

(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7

8 0
3 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
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