What is the volume of solution needed to make a 0.5M solution, using 250g of NaCl

Answers

Answer 1
Answer:

1 mole of NaCl has (23+35.5)g= 58.5g

Therefore, to  make 1M/L of solution you will need 58.5g Then to make 0.5M/L you will need 58.5/2=29.25g

If to make 1liter of 0.5M one will need 29.25 grams of NaCl, then250g of NaCl will make

(250g×1L)/29.25g= 8.547liters


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Describe the specific location where the head of the caudate nucleus is continuous with its body within the brain. What is the significance of this continuity, and how does it relate to the functions associated with the caudate nucleus?

Answers

Final Answer:

The head of the caudate nucleus blends with its body along the lateral wall of the lateral ventricles. This continuity enables integration with neighboring brain regions, crucial for motor, cognitive, and emotional functions.

Explanation:

The caudate nucleus, a C-shaped structure within the brain, consists of a head, body, and tail. The head of the caudate nucleus, situated at the anterior end, smoothly transitions into its body. This junction occurs along the lateral wall of the lateral ventricles, which are fluid-filled cavities that play a role in cerebrospinal fluid circulation.

The significance of this continuity lies in the structural integration it provides. This seamless connection allows the caudate nucleus to interact with neighboring brain regions, such as the putamen, globus pallidus, and cerebral cortex. These connections form part of the basal ganglia, a complex network involved in motor control, cognition, and emotional regulation.

Through these intricate interconnections, the caudate nucleus contributes to various functions. It plays a pivotal role in motor planning and execution, facilitating smooth, coordinated movements. Additionally, it participates in cognitive processes, including learning, memory, and decision-making. Moreover, the caudate nucleus is implicated in emotional regulation, influencing aspects of motivation and reward processing.

In summary, the seamless continuity between the head and body of the caudate nucleus along the lateral ventricles is essential for its functional integration with neighboring brain regions. This integration underpins the caudate nucleus's involvement in motor, cognitive, and emotional processes, highlighting its significance in overall brain function.

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Explanation:

The head of the caudate nucleus is continuous with its body at a specific location called the genu of the internal capsule. The genu of the internal capsule is a bend or curve in the white matter fibers that connect different regions of the brain.

The significance of this continuity between the head and body of the caudate nucleus at the genu of the internal capsule lies in the anatomical connectivity it provides. The genu of the internal capsule serves as a major pathway for communication between the cortex and the subcortical structures, including the caudate nucleus.

This continuity allows for bidirectional communication between the cortex and the caudate nucleus, facilitating the integration of information related to motor control, cognition, and emotion. It enables the caudate nucleus to receive input from various cortical regions, especially the prefrontal cortex, which is involved in executive functions and decision-making.

The caudate nucleus, including its head and body, is implicated in various functions such as motor control, learning, memory, reward processing, and cognitive flexibility. The continuity at the genu of the internal capsule ensures that signals from the cortex can be transmitted to the caudate nucleus, allowing it to integrate and modulate these functions.

Furthermore, disruptions or lesions at the genu of the internal capsule can lead to motor and cognitive impairments, as the communication between the cortex and the caudate nucleus gets compromised. This highlights the importance of the continuity between the head and body of the caudate nucleus at this specific location for the proper functioning of the brain and its associated functions.

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How many grams are in 2.3 x 10^-4 moles of Ca3(PO4)2

Answers

Molar mass 

Ca₃(PO₄)₂ = 310 g/mol

1 mole -------------------> 310 g
2.3x10⁻⁴ mole ---------> ?

m = 2.3x10⁻⁴ * 310 / 1

m = 0.0713 g 

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A sample of seawater has a mass of 158g and a volume of 156mL. What is its density?

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A sample of seawater has a mass of 158g and a volume of 156mL. its density is 1.012 g/mL.

According to the question the data are given is as follows:

volume of sea water  = 156 mL

mass of sea water = 158 g

density of sea water can be calculated by the following formula :

D = M / V

where,

D = density

M = mass of sample

V = volume of sample

substituting all the value in the given formula , we get :

D = 158 g / 156 mL = 1.012 g/mL

Density of seawater = 1.012 g / mL

Thus, A sample of seawater has a mass of 158g and a volume of 156mL. its density is 1.012 g/mL .

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You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2.00 *10^4 Kpa at 28c How many kilograms of N2 does the cylinder contain

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The quantity of nitrogen the cylinder contains is 4477.8 g

What is pressure?

Pressure is a force exerted in a perpendicular direction in any item.

By ideal gas law

PV = nRT

P =(w)/(M) =(RT)/(V)\n

w = mass

Volume is 20.0 l

Pressure is  2.00 * 10^4 \;Kpa

The molar mass of nitrogen is 28 g/mol

R is gas constant = 0.0821

Temperature is 28 converted into kelvin that is 301 k

Putting the values

197.6 =(w)/(28) =(0.0821 * 301 )/(20\;l)\n\nw = 4477.8 g

Thus, the mass of nitrogen is 4477.8 g.

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Answer : The mass of N_2 gas is, 4477.8 g

Solution :

using ideal gas equation,

PV=nRT\n\nP=(w)/(M)* (RT)/(V)

where,

n = number of moles of gas

w = mass of gas

P = pressure of the gas = 2* 10^4Kpa=197.6atm

conversion :1atm=101.2Kpa

T = temperature of the gas = 28^oC=273+28=301K

M = molar mass of N_2 gas = 28 g/mole

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = 20 L

Now put all the given values in the above equation, we get the mass of gas.

197.6atm=(w)/(28g/mole)* (0.0821Latm/moleK* 301K)/(20L)

w=4477.8g

Therefore, the mass of N_2 gas is, 4477.8 g

The gram formula mass of NH4Cl is(1) 22.4 g/mole (3) 53.5 g/mole(2) 28.0 g/mole (4) 95.5 g/mole

Answers

Answer: Option (3) is the correct answer.

Explanation:

Gram formula mass or molar mass is the sum of mass of all the individual  atoms present in the compound or formula.

For example, gram formula mass of NH_(4)Cl is calculated as follows.

     NH_(4)Cl = Mass of N + 4 * mass of H + mass of Cl

                            = (14.00 + 4 * 1.00 + 35.45) g/mol

                             = (14.00 + 4.00 + 35.45) g/mol

                             = 53.45 g/mol

                             = 53.5 g/mol (approx)

Thus, we can conclude that the gram formula mass of NH_(4)Cl is 53.5 g/mol (approx)

The gram formula mass of NH₄Cl is 53.5 g/mol (3rd option)

How to determine the gram formula mass?

The gram formula mass of a compound is simply the molar mass of the compound.

The gram formula mass of NH₄Cl can be obtained as illustrated below:

  • Molar mass of N = 14 g/mol
  • Molar mass of H = 1 g/mol
  • Molar mass of Cl = 35.5 g/mol
  • Gram formula mass of NH₄Cl = ?

Gram formula mass of NH₄Cl= 14 + (4 × 1) + 35.5

= 14 + 4 + 35.5

= 53.5 g/mol

Thus, we can conclude that the gram formula mass of NH₄Cl is 53.5 g/mol (3rd option)

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A substance conducts an electric current when it is in the liquid state. What would the conductivity of the substance in solid form indicate about the type of bond in the substance?A.) If the solid is conductive, the bonds are ionic.
B.) If the solid is not conductive, the bonds are covalent.
C.) If the solid is conductive, the bonds are metallic.
D.) If the solid is not conductive, the bonds are metallic.

Answers

Answer is: C.) If the solid is conductive, the bonds are metallic.

1) Substances with ionic bond conduct an electric current in liquid, but not in solid state, because in liquids ions are mobile, contrary to solids where ions are fixed.

2) Substances with covalent bond not conduct an electric current in liquid and solid state, because they not have free ions or electrons.

3) Substances with metallic  bond conduct an electric current in liquid and solid state, because they have mobile electrons. Most metals have strong metallic bond, because strong electrostatic attractive force between valence electrons (metals usually have low ionization energy and lose electrons easy) and positively charged metal ions.

Answer:C.) If the solid is conductive, the bonds are metallic.

Explanation: