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2 years ago

You live 6 kilometers from your school. How many meters do you live from school?

Chemistry

2 answers:

Phantasy [73]2 years ago

7 0

1000 m = 1 km
so if you have 6 km you have 1000×6= 6000m
So you live 6000m from school

Snezhnost [94]2 years ago

5 0

Answer: 6,000 m . (or, write as: 6,000 meters).
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Explanation:
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Note the exact conversion: 1 km = 1000 m .
______________________________________________________
Note: "km" is an abbreviated form of "kilometer(s)".
"m" is an abbreviated form of "meter(s)".
______________________________________________________
Here is one way to solve, using a method called "dimensional analysis":
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Given: 6 km ; convert to "m" ;
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→ 6 km = ? m ;
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 → 6 km * $\frac{1000 m}{1 km}$ = ? m ;
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→ On the "left-hand side" of the equation, the units of "km" "cancel out" to "1" ; and we are left with:
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→ 6 * $\frac{1000 m}{1$ = ? m ;
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.→ 6 * 1000 m = ? m ;
_______________________________________________
→ = 6,000 m ; or write as: 6,000 meters .
___________________________________________________

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Limiting reactants would appreciate the help

Vanyuwa [196]

Answer:

Explanation:

The Limiting Reactant is that reactant which when consumed in a reaction stops the reaction. The other reactants will be in excess and typically considered non-reactive.

To identify the limiting reactant ...

- write and balance the reaction of interest. Express it in standard form. That is, standard form of a reaction is when the coefficients of the balanced equation are in their lowest whole number values. Also, remember that the standard equation is 'assumed' to be at STP conditions (0°C & 1atm).

- convert all given reactant values to moles

- divide each reactant mole value by the related coefficient of the the balanced standard equation. The smaller value is the limiting reactant. The remaining reactants will be in excess.

Your Problem:

Given: 3Ba + N₂ => Ba₃N₂

22.6g 4.2g ?

moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba

moles N₂ => 4.2g/14.007g/mol= 0.150 mole N₂

Part A: Determining the Limited Reactant

Divide each mole value by respective coefficient ... smallest value is Limiting Reactant.

Barium => 0.165/3 = 0.055 <=> (Limiting Reactant)

Nitrogen => 0.15/1 = 0.15

Barium is the smaller result and is therefore the limiting reactant. This works for ALL limiting reactant type problems. However, be sure to use the mole values calculated first (Ba = 0.165mol & N₂ = 0.150mol) when doing ratio calculations.

Part B: Max (theoretical) amount of Ba₃N₂ produced:

Note: The product yield amounts are based upon the given 'moles' of limiting reactant, NOT the results of the 'divide by respective coefficient' step used to ID the limiting reactant.

3Ba + N₂ => Ba₃N₂ (3:1 rxn ratio for Ba:Ba₃N₂)

moles 0.165mole 0.150mole 1/3(0.165)mole = 0.055mole Ba₃N₂

= 0.055mol(440g/mol) Ba₃N₂

= 24.2 grams Ba₃N₂ (as based

upon Barium as Limiting Reactant)

Part C: Excess N₂ remaining after reaction stops:

From balanced standard reaction, the reaction ratio for Ba:N₂ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) = moles of N₂ used.

moles of N₂ used = 1/3(0.165)mole = 0.055mole N₂ used

∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.

mass N₂ remaining = 0.095mole x 28g/mole = 2.66 grams N₂ remaining in excess.

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77julia77 [94]

Answer:

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Explanation: Hope this helps!

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2 years ago

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snow_tiger [21]

Answer:

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