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Natalka [10]
3 years ago
10

Nickel (iii) cyanide + aluminum permanganate product?

Chemistry
1 answer:
Wewaii [24]3 years ago
5 0

Answer:

Nickel(II) cyanide is an inorganic compound with a chemical formula Ni(CN)₂

Explanation:

<em>Hope </em><em>it </em><em>helps </em><em>u </em>

FOLLOW MY ACCOUNT PLS PLS

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One liter of a buffer composed of 1.2 m hno2 and 0.8 m nano2 is mixed with 400 ml of 0.5 m naoh. what is the new ph? assume the
enyata [817]
Answer is: 3,4
Chemical reaction: HNO₂ + NaOH → NaNO₂ + H₂O.
c₀(HNO₂) = 1,2 M = 1,2 mol/dm³.
c₀(NaNO₂) = 0,8 M = 0,8 mol/dm³.
V₀(HNO₂) = V₀(NaNO₂)  = 1 dm³ = 1 L.
c₀(NaOH) = 0,5 M = 0,5 mol/dm³.
n₀(HNO₂)= 1,2 mol/dm³ · 1 dm³ = 1,2 mol.
n₀(NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.
V(NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.
n₀(NaOH) = c₀(NaOH) · V₀(NaOH).
n₀(NaOH) = 0,5 mol/dm³ · 0,4 dm³ = 0,2 mol.
n(HNO₂) = 1,2 mol - 0,2 mol = 1 mol.
n(NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.
c(HNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
c(NaNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
pH = pKa + log (c(HNO₂) / c(NaNO₂)).
pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.
6 0
3 years ago
Read 2 more answers
Grace measures 20mL of water in a graduated cylinder. she puts a piece of rock in the
GaryK [48]

The volume would be explanation times explanation because the explanation is the explanation

Explanation:

3 0
3 years ago
Read 2 more answers
What happened when Crookes placed a solid object inside
telo118 [61]
The answer I believe would be C hope this helps
7 0
3 years ago
THIS IS FOR A QUIZ DUE IN 5 MINS
RideAnS [48]

Answer:

12.18 u

Explanation:

The average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its stable isotopes.

5 0
3 years ago
Convert 9.32x 10 23ª atoms of au to moles of au
7nadin3 [17]

Answer:

\huge\boxed{\sf no.\ of\ moles = 1.55\ moles }

Explanation:

<u>Given:</u>

Number of atoms = 9.32 \times 10^{23} atoms

Avogadro's Number = 6.023 \times 10^{23} atom / mol

<u>Required:</u>

Moles = ?

<u>Formula:</u>

\displaystyle No.\ of\ moles = \frac{no. \ of \ atoms }{avogadro's \ no.}

<u>Solution:</u>

\displaystyle no. \ of \ moles = \frac{9.32\times 10^{23}}{6.023 \times 10^{23}}

no. of moles = 1.55 moles

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
5 0
3 years ago
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