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andreyandreev [35.5K]
2 years ago
14

Which of the following best describes a single replacement reaction?

Chemistry
1 answer:
Lady_Fox [76]2 years ago
6 0

Answer:

One element takes the place of another in a compound

Explanation:

I just took a test for it and got it right. :)

Hope This Helps :)

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D because these are the biggest particles, and are therefore, the most dense. 
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professor190 [17]

Answer:

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Explanation:

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In addition to the two types of motion found in liquids, gases have <br> What kind of motion
makvit [3.9K]

Answer:

In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles.

3 0
3 years ago
3. Hydrogen reacts with nitrogen to produce ammonia according to the equation:
lys-0071 [83]

Mass of ammonia produced : 121.38 g

<h3>Further explanation</h3>

Given

Reaction

3H₂(g) + N₂(g) ⇒ 2NH₃(g)

100g of N₂

Required

Ammonia produced

Solution

mol of N₂ :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{100}{28}\\\\mol=3.57

From the equation, mol ratio of N₂ and NH₃ = 1 : 2, so mol NH₃ :

\tt \dfrac{2}{1}\times 3.57=7.14~moles

mass of NH₃(MW=17 g/mol) :

\tt mass=mol\times MW\\\\mass=7.14\times 17\\\\mass=121.38~g

8 0
3 years ago
If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio
kati45 [8]

Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>
8 0
2 years ago
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