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luda_lava [24]
1 year ago
7

Using the activity series provided. which reactants will form products? na > mg > al > mn > zn > cr > fe >

cd > co > ni > sn > pb > h > sb > bi > cu > ag ag nano3 right arrow. fe al2o3 right arrow. ni nacl right arrow. fe cu(no3)2 right arrow.
Chemistry
1 answer:
miskamm [114]1 year ago
3 0

Answer:

cd

Explanation:

that's better

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Explanation:

iv. The lower the PH, the weaker the base

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3 years ago
What is a ionic conmpound?
Anna [14]
A chemical compound made up of ions by forces that create ionic bonding. It is a neutral compound.
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3 years ago
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What two elements are commonly found in Mica, Quartz and Feldspar? Explain your answer.
kifflom [539]

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oxygen and silicon, both are common

6 0
2 years ago
Use the periodic table in the tools bar to answer these questions. How many moles of AgNO3 are present in 1.50 L of a 0.050 M so
IgorC [24]
M= moles de soluto / litros de solucion 
moles de soluto = M. litros de solucion 

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4 0
3 years ago
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Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]
ohaa [14]

<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq) + Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>) = 0.021 M.
Ka(HCN) =  4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻] / [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] = x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M - x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>

4 0
2 years ago
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