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ozzi
3 years ago
13

How many grams of KNO3, are needed to create 675.0 mL with a concentration of 1.71 M

Chemistry
1 answer:
valkas [14]3 years ago
4 0

Answer:

Бардык белгилүү авиация мыйзамдарына ылайык, аары учуу мүмкүнчүлүгү жок. Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.

Explanation:

Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.

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26.8 moles of argon contain how many atoms?
stepladder [879]

Answer:

the answers is A

Explanation:

4 0
4 years ago
About how many centimeters will make an inch?<br> A)5<br> B)2+<br> C) 200<br> D)10
Rasek [7]

Answer: B) 2+

Explanation:

   There are exactly 2.54 centimeters in an inch, the closest option is:

B) 2+

6 0
2 years ago
Read 2 more answers
Water and oxygen gas are the products of a chemical reaction.
Darina [25.2K]
The answer is C.

H₂O₂ ----> H₂O + O₂
8 0
3 years ago
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Manganese metal can be obtained by the reaction of manganese dioxide with aluminium. 4Al(s) + 3MnOz(s) → 2 Al2O3(s) + 3 Mn(s) Ca
Snezhnost [94]

Answer : The enthalpy change of reaction is -1800 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given final reaction is,

4Al(s)+3MnO_2(s)\rightarrow 2Al_2O_3(s)+3Mn(s)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2Al(s)+\frac{3}{2}O_2(g)\rightarrow Al_2O_3(s)    \Delta H_1=-1680kJ/mole

(2) Mn(s)+O_2(g)\rightarrow MnO_2(s)    \Delta H_2=-520kJ/mole

First we will multiply reaction 1  by 2 and reverse reaction of reaction 2 by 3 then adding both the equation, we get :

The expression for final enthalpy is,

\Delta H=[n\times \Delta H_1]+[n\times (-\Delta H_2)]

where,

n = number of moles

\Delta H=[2mole\times (-1680kJ/mole)]+[3\times -(-520kJ/mole)]

\Delta H=-1800kJ

Therefore, the enthalpy change of reaction is -1800 kJ

8 0
3 years ago
How many moles FeBr3 are required<br> to generate 275 g NaBr?<br><br> 2FeBr3 + 3Na₂S → Fe₂S3 +6NaBr
slega [8]

Answer:

0.893mol

Explanation:

n = m ÷ M

= 275 ÷ (23 + 80)

= 2.67 mol

* now use the Mol ratio *

NaBr : FeBr3

6 : 2

2.67 : x

5.67 = 6x

n( FeBr3 ) = 0.893 mol

8 0
2 years ago
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