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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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A waste treatment settling tank treats an industrial waste inflow (QIN) of 0.2 m3/s with a suspended particulate concentration (
iren2701 [21]

The effluent flow in concentration and particulate mass flow will be 0.198m³/sec.

<h3>How to calculate the effluent flow?</h3>

It should be noted that the total inflow will be equal to the total outflow. Therefore,

0.2 + 0.048 = 0.05 + We

Collect like terms

Qe = 0.2 + 0.048 - 0.05

Qe = 0.198m³/sec

The concentration will be:

= (360 × 1000)/0.05

= 7200mg/L.

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8 0
2 years ago
A 20.0 milliliter sample of 0.200 molar K₂CO₃ solution is added to 30.0 milliliters of 0.400 molar Ba(NO₃)₂ solution. Barium car
sveticcg [70]

Answer:

0.32M

Explanation:

<u>Step 1:</u>  Balance the reaction

K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3

We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution

SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3

<u>Step 2: </u>Calculate concentration

To find the concentration of the barium cation we use the following equation:

Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>

<u />

<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3

[Ba2+] = 0.32 M

The concentration of Barium ion in solution is 0.32 M

6 0
3 years ago
A prototype is usually different from the final product
Assoli18 [71]
True. The prototype is usually the "rough draft" the figure out what needs fixed or upgraded before they make the final product "final draft". Hope that helped!
7 0
3 years ago
PLEASE HELP 20 POINTS What is a key event found in CO2 and glucose?
irina [24]

the answer is carbon

5 0
3 years ago
Al + AgNO3 -&gt; Al(NO3)3 + Ag. As a balanced equation​
prisoha [69]

Answer:

Al + 4AgNO3 >>Al(NO3)3+ 3Ag

Explanation:

the number of moles of No3 of the products is 3 therefore we have to balance the reactants by adding 3 before the "AgNO3" which also leades us to adding 3 mols to Ag on the products side

4 0
2 years ago
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