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Eva8 [605]
3 years ago
7

B) Which of the following is derived quantity?i. Lengthii. Temperatureiii. Density​

Chemistry
1 answer:
AnnyKZ [126]3 years ago
7 0

Answer:

3,density

Explanation:

because it doesn't measure dirrectly

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How many ions are there in 0.187 mol of Na+ ions
kondaur [170]
Best Answer
1 mole of a substance contains 6.022x10^23 "units" of that substance.

So 0.187 mol of Na+ is 1.13x10^23 ions (6.022x10^23 x 0.187).
4 0
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How is the AHfusion used to calculate volume of liquid frozen that produces 1
yulyashka [42]

Answer:

option B

Explanation:

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3 years ago
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C55H72MgN4O5. How many atoms are in this molecule?
Bas_tet [7]
There are 137 atoms in this molecule. C55 + H72 = 127. 127 + Mg (one atom of magnesium = 128. 128 + N4 = 132. 132 + O5 = 137.
5 0
3 years ago
If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
Alina [70]

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

3 0
3 years ago
0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction
algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce \frac{2}{1} = \frac{x}{0.50}

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

\frac{2}{2} = \frac{x}{0.6}

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

\frac{2}{3} = \frac{x}{0.9}

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0
3 years ago
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