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3 years ago

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Which has a greater speed,a heron that travels 600m in 60 seconds or a duck that travels 60m in 5 seconds? explain

1 answer:

Semmy [17]3 years ago

6 0

In order to find out the answer to this problem, you need to know how to find the unit rate. A unit rate a rate that has a denominator of 1. You find the unit rate because the problem is asking you to find the faster speed, but the ratios are different.
So 600m in 60 sec you divide both by 60. It comes out as 10m in 1 sec for the heron.
60m in 5 sec is 12m in 1 sec for the duck.
Compare the heron to duck. The heron travels faster.
Give 5 stars if it helped!! :)

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The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kj . f the change in e

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<u>Given:</u>

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

<u>To determine:</u>

The work done, W

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Based on the first law of thermodynamics,

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the work done by a gas is given as:

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W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

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What is the empirical formula of a compound that is made of 3.80g of phosphorus and 0.371g of hydrogen?

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3 years ago

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

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