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ale4655 [162]
3 years ago
10

Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?

Chemistry
1 answer:
TiliK225 [7]3 years ago
7 0

Answer:- HBr is limiting reactant.

Solution:- The given balanced equation is:

2Al+6HBr\rightarrow 2AlBr_3+3H_2

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

8molAl(\frac{6molHBr}{2molAl})

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

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I need help with these two
bearhunter [10]

Answer:

6. d,  7. a

Explanation:

6. Molarity is a number of moles solute in 1 L solution.

7. 1 L solution - 2.5 mol K2CO3

 20 L             - x mol K2CO3

x  =20*2.5/1 = 50 mol K2CO3

Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol

99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.

Actually KCO3 does not exist, in reality it should be K2CO3.

3 0
3 years ago
A 50.0-g sample of liquid water at 25.0 °c is mixed with 23.0 g of water at 79.0 °c. the final temperature of the water is ___
I am Lyosha [343]

<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>

<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>

<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>

<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>

<span>73.0<span>Tf</span>=2561 °C</span>

<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>

8 0
3 years ago
What best describes how the behavior of rocks changes when they become deeply buried and placed under high levels of heat and pr
Debora [2.8K]

Answer:

They become ductile and deform plastically

Explanation:

When rocks are buried by the materials up to a greater depth, then the confining pressure increases significantly. This results in the ductile behavior of the rocks at such depth. These rocks are present in the ductile region where the depth is about more than 20 to 30 km. Here the rocks are subjected to extremely high pressure and temperature conditions, which favors the transformation of rocks into more higher-grade metamorphic rocks. It is also enhanced due to the geothermal gradient.

Under such high pressure and temperature, the rocks show the behavior of plasticity, where the rocks undergo bending, buckling as well as they tend to flow, and there occurs low strain rate, resulting in the permanent deformation of rocks.

Thus, the rocks become ductile and deform plastically at such conditions.

5 0
3 years ago
If a substance has a mass of 12.50 g and takes up 3.4 mL of space, what is the density?
miskamm [114]

Answer:

The answer is

<h2>3.68 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>density =  \frac{mass}{volume}</h3>

From the question

mass of substance = 12.50 g

volume = 3.4 mL

The density of the substance is

density =  \frac{12.50}{3.4}  \\  = 3.676470588...

We have the final answer as

<h3>3.68 g/mL</h3>

Hope this helps you

7 0
3 years ago
17
leonid [27]

Answer:

Which of the following

properties distinguishes a solution

oversaturated with a dilute?

  • The supersaturated solution is one in which the solvent has dissolved more solute than it can dissolve in the saturation equilibrium.  The solute can be a solid, or a gas. The molecules of the solvent surround those of the solute and seek to open space between themselves to be able to harbor more amount of solute.
  • A dilute solution is a solution that has not reached the maximum concentration of solute dissolved in a solvent. The additional solute will dissolve when added in a dilute solution and will not appear in the aqueous phase.   It is considered a state of dynamic equilibrium where the speeds in which the solvent dissolves the solute are greater than the recrystallization rate.
8 0
3 years ago
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