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insens350 [35]
3 years ago
14

4.8 tons of pig ovaries were required to extract 12.0 mg of estrogen

Chemistry
1 answer:
bearhunter [10]3 years ago
7 0
Is this a question? if so please elaborate.
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What is earths spheres
Oliga [24]
Geosphere >> Which is Earth's entire solid body..

Hydrosphere >> Which is all the "liquid" water on Earth..

Atmosphere >> Which is the few layers of "gasses" that are surrounding Earth..

Cryosphere >> Which is all the "frozen" water on Earth..

Biosphere >> Us! All the living organisms on Earth..


6 0
4 years ago
Read 2 more answers
Need help with balancing equations
solniwko [45]

The balanced equation of the reaction shows that the mole of atoms on both sides of the reaction are equal.

<h3>What are balanced chemical equations?</h3>

Balanced chemical equations are equations in wgich the moles of each species on both sides of the reaction are equal.

To balance chemical equations, numerical coefficients are placed in front of the species in the reaction.

The balanced equations of the reaction are as follows:

3Pd + 4H_3PO_4 \rightarrow Pd_3(PO_4)_4 + 6H_2 \\

C_{11}H_{24} + 17O_2 \rightarrow 11CO_2 + 12H_2O \\

2Al(HCO_3)_3 + 3H_2SO_4\rightarrow Al_2(SO_4)_3 +6CO_2 +6H_2O \\

Therefore, the balanced equation of the reaction shows that the mole of atoms on both sides of the reaction are equal.

Learn more about balancing equations at: brainly.com/question/11904811

#SPJ1

7 0
2 years ago
Which term is used to describe all the habitats around the world?
OlgaM077 [116]

Answer:

ecosystem diversity

Explanation:

hope this helped

5 0
2 years ago
True or false stars will eventually burn out causing the universe become cold and dark
Art [367]

Answer:

False

Explanation:

At the end of the 1000 years, Satan will be released, defeated again, and then cast into the lake of fire (Revelation 20:7-10). Then, after a final judgment by God, the end of the world described in 2 Peter 3:10 occurs. The Bible tells us several things about this event.

7 0
3 years ago
Read 2 more answers
What is the molecular mass for a non-electrolyte if 35.0 g of it is dissolved in 45.0 grams of water and the solutions boiling p
adelina 88 [10]

The boiling point of water increases as the amount of impurities dissolved in it increases. For our purposes, we will consider the non-electrolyte to be the dissolved impurity. The change in the boiling point can be calculated using the equation:

\Delta T_b = i \times K_b \times m

where \Delta T_b is the change in boiling point, i is the van ‘t Hoff factor (whose value denotes the number of particles each formula unit of the dissolved substance dissociates into in water), K_b is the boiling point elevation constant, and m is the molality (moles of solute/kilogram of solvent) of the solution.

Right off the bat, since we're dealing with a non-electrolyte, the dissolved substance can be assumed not to dissociate in water. So, our van ‘t Hoff factor, i, would be 1 (by contrast, the i for an ionic compound like NaCl would be 2 since, in water, NaCl would dissociate into two particles: one Na⁺ ion and one Cl⁻ ion). We're also given our K_b, which is 0.51 °C/<em>m</em>.

Assuming the normal boiling point of pure water to be 100 °C (a defined value for sig fig purposes), the change in boiling point from having dissolved 35.0 g of the non-electrolyte can be obtained by subtracting 100 °C from the final—elevated—boiling point of 101.25 °C:

\Delta T_b = 101.25\text{ }^o\text{C} - 100\text{ }^o\text{C} = 1.25\text{ }^o\text{C}

Now, recall what we're asked to determine: the molecular mass of the dissolved substance. There is one unknown left in the equation: the molality of the solution. Let's first solve for that:

m = \frac{\Delta T_b}{K_b} = \frac{1.25^\text{ o}\text{C}}{0.51^\text{ o}\text{C}/m} \\ m = 2.45 \text{ mol solute/kg water}.

Notice that we didn't include the <em>i </em>since its value is 1.

Now, what would happen if we multiplied our molality by the mass of water we've been given? We would be left with the moles of solute. And what are we asked to find? The molecular mass, or the mass per mole. We can accomplish this in two steps. Remember to convert your mass of water to kilograms:

2.45 \text{ mol solute/kg water} \times 0.045 \text{ kg water} = 0.110 \text{ mol solute.}

And, finally, we divide the mass of our solute by the number of moles of solute:

\frac{35.0 \text{ g solute}}{0.110 \text{ mol solute}} = 317.5 \text{ g/mol}

Our answer to two significant figures (which is the number of sig figs to which our K_b is given) would be 320 g/mol.

5 0
3 years ago
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