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antiseptic1488 [7]
3 years ago
5

The most important part of the brain for an organism is the...​

Chemistry
1 answer:
Serjik [45]3 years ago
4 0

Answer:

its the lower brainstem

Explanation:

The lowest part of the brainstem, the medulla is the most vital part of the entire brain and contains important control centers for the heart and lungs.

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I SWEAR THIS IS WAY TOO EASY, WHY AM I THAT STUPID, HELP ME AND ILL MARK U BRAINLIEST
exis [7]

Answer:

i think the first one is gravity and second one is rotation

Explanation:

5 0
3 years ago
Hcl and nh3 react to form a white solid, nh4cl. if cotton plugs saturated with aqueous solutions of each are placed at the ends
IgorLugansk [536]

24.4 cm.

<h3>Explanation</h3>

HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.

The speed of a HCl or NH₃ molecule depends on its <em>kinetic energy</em>.

E_\text{k} = 1/2 \; m \cdot v^{2}

Where

  • E_\text{k} is the <em>kinetic energy</em> of the molecule,
  • m its mass, and
  • v^{2} the square of its speed.

Besides, the <em>kinetic theory</em> <em>of gases</em> suggests that for an ideal gas,

E_\text{k} \propto T

where \text{T} its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the <em>average kinetic energy</em> of molecules shall be the same for <em>any ideal gas </em>at the same<em> temperature</em>. So is the case for HCl and NH₃

E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)

m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)

Where

  • m(\text{HCl}), v(\text{HCl}), and E_\text{k}(\text{NH_3}) the mass, speed, and kinetic energy of an HCl molecule;
  • m(\text{NH}_3), v(\text{NH}_3), and E_\text{k}(\text{NH}_3) the mass, speed, and kinetic energy of a NH₃ molecule.

The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their <em>molar mass</em>. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, m(\text{HCl}) = 36.45 / 17.03 \; m(\text{NH}_3). Therefore:

36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)

36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)

\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)

The <em>average </em>speed NH₃ molecules would be  \sqrt{36.45/17.03} \approx 1.463 <em>if</em>  the <em>average </em>speed of HCl molecules v(\text{HCl}) is 1.

\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}

\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}

8 0
3 years ago
If 50 milliliters of a 1.0M NaOH solution is
lorasvet [3.4K]

Answer:

c

Explanation:

6 0
3 years ago
Phosphorite is a mineral that contains plus other non-phosphorus-containing compounds. What is the maximum amount of that can be
8_murik_8 [283]

Phosphorus can be prepared from calcium phosphate by the following reaction:

2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)

Phosphorite is a mineral that contains Ca_3(PO_4)_2 plus other non-phosphorus-containing compounds. What is the maximum amount of P_4 that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75% Ca_3(PO_4)_2 by mass? Assume an excess of the other reactants.

Answer: Thus the maximum amount of P_4 that can be produced is 0.345 kg

Explanation:

Given mass of phosphorite Ca_3(PO_4)_2  = 2.3 kg

As given percentage of phosphorite Ca_3(PO_4)_2 is = \frac{75}{100}\times 2.3kg=1.725kg=1725g

moles=\frac{\text {given mass}}{\text {Molar mass}}

{\text {moles of}Ca_3(PO_4)_2=\frac{1725g}{310g/mol}=5.56moles

2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)

According to stoichiometry:

2 moles of phosphorite gives = 1 mole of P_4

Thus 5.56 moles of phosphorite give= \frac{1}{2}\times 5.56=2.78moles of P_4

Mass of P_4=moles\times {\text {Molar mass}}=2.78mol\times 124g/mol=345g=0.345kg

Thus the maximum amount of P_4 that can be produced is 0.345 kg

5 0
3 years ago
Given that the density of water is 0.975 g/mL and that 171 g of sucrose (molar mass: 342.30 g/mol) is dissolved in 512.85 mL of
nikitadnepr [17]
1.00 m is the molality of this solution
7 0
3 years ago
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