CHEMISTRY HIGH SCHOOL

What location is 90 degrees away from the equator?A) The south pole
B) Polaris
C) Both A & B
D) The north pole

Answers

Answer 1
Answer: C) both A and B

Each parallel measures one degree north or south of the Equator, with 90 degrees north of the Equator and 90 degrees south of the Equator
Hope this helps!
Answer 2
Answer:

Answer:

C) Both A and B

Explanation:

The axis of 0 degrees latitude is known as the Equator. With 90 degrees north of the Equator and 90 degrees south of the Equator, each parallel measures one degree north or south of the Equator. The North Pole is 90 degrees north latitude, and the South Pole is 90 degrees south latitude.

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44 points! Will medal!3. A calorimeter contains 500 g of water at 25°C. You place a hand warmer containing 200 g of liquid sodium acetate inside the calorimeter. When the sodium acetate finishes crystallizing, the temperature of the water inside the calorimeter is 39.4°C. The specific heat of water is 4.18 J/g-°C. How much energy was needed to heat the water?

Answers

A calorimeter contains 500 g of water at 25°C.....

the temperature of the water inside the calorimeter is 39.4°C.....

The specific heat of water is 4.18 J/g-°C.

energy needed to heat the water = specific heat * mass * temp difference

= 4.18 J/g-°C * 500 g * (39.4°C - 25°C)

= 4.18*500*14.4

= 30096J

or approx. 30kJ

Energy=specific heat of water x mass of water x water's temperatures

=4.18x500x(39.4-25)

=30096J

sodium chloride and glucose both are soluble in water but the solubility of NaCl is greater then glucose why.

Answers

The solubility of a substance in water depends on several factors, including the nature of the solute and solvent, as well as the intermolecular forces involved.

In the case of sodium chloride (NaCl) and glucose, both substances are indeed soluble in water. However, the solubility of NaCl is generally greater than that of glucose due to differences in their chemical properties.

NaCl is an ionic compound, meaning it consists of positive sodium ions (Na+) and negative chloride ions (Cl-). When NaCl is added to water, the polar water molecules surround the ions and separate them from the crystal lattice. These water molecules form favorable interactions with the charged ions through ion-dipole attractions, resulting in the dissolution of NaCl in water. These strong ion-dipole forces contribute to the high solubility of NaCl in water.

On the other hand, glucose is a covalent compound composed of carbon, hydrogen, and oxygen atoms. It does not dissociate into ions when dissolved in water. Instead, glucose molecules interact with water through weaker intermolecular forces such as hydrogen bonding and dipole-dipole interactions. While these forces allow glucose to dissolve in water, they are relatively weaker than the ion-dipole interactions in NaCl-water solution. As a result, glucose has a lower solubility compared to NaCl in water.

Therefore, the differing chemical properties of sodium chloride and glucose contribute to the higher solubility of NaCl in water compared to glucose.



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We can also use the equation for enthalpy change for physical phase changes. Consider the phase change H2O(l) → H2O(g). Calculate ΔHrxn.Use ΔHf values from the table above.

ΔHrxn = _____

Answers

Answer: The value of enthalpy of the given reaction is 44.000 kJ/mol , that is energy is supplied to water to change into water vapors

Explanation:

H_2O(l)\rightarrow H_2O(g),\Delta H_(rxn)=?

Enthalpy of formation of water in liquid state,\Delta H_(f,H_2O(l))=-285.820 kJ/mol

Enthalpy of formation of water in gaseous state,\Delta H_(f,H_2O(g))=-241.820 kJ/mol

\Delta H_(rxn)=\sum \Delta H_f \text{of products}-\sum \Delta H_f \text{of reactants}

\Delta H_(rxn)=(-241.820 kJ/mol)-(-285.820 kJ/mol)=44.000 kJ/mol

The value of enthalpy of the given reaction is 44.000 kJ/mol, that is energy is supplied to water to change into water vapors.
Standard molar enthalpy:
H2O ( liquid water ) : - 285.8 KJ/mole
H2O ( water vapor ) : - 241.8 KJ / mole.
ΔHrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0
Answer: B ) 44.0 KJ  

ovalent bonds are formed by which of the following? A. The energy transfer from one body to another B. The bonding of cations and anions C. Weak electrostatic attraction between charged areas of adjacent molecules D. The electrostatic attraction between oppositely charged ions E. The sharing of electrons between atoms

Answers

Answer:

E. THE SHARING OF ELECTRONS BETWEEN ATOMS

Explanation:

Covalent bonds are formed by sharing of electrons between the two reacting atoms so that a stable octet structure can be attained. The electron pair is called shared pair and each atom contributes an electron of the shared pair of electron.

Molecules are formed by covalent bonding and not ions. Examples of covalent bonds are hydrogen molecule (H2), Chlorine molecule (Cl2), water molecule (H2O).

Covalent compounds usually dissolve in non-polar solvents and not polar solvents (water), most covalent compounds do not conduct electricity.

Dative bonding is different from normal covalent bindings as the shared pair is donated by only one of the participating atom in coordinate covalent bonding.

When iron metal reacts with oxygen, the reaction can form Fe2O3. Write a balanced chemical equation for this reaction, and find the number of moles of oxygen that are needed to form 6 mol of Fe2O3 while showing your work.

Answers

The balanced equation is 
4Fe+3O₂⇒2Fe₂O₃

We know that the mole of Fe₂O₃ is 6, and since the ratio between oxygen and Fe₂O₃ is 3:2, we can see that

3:2 = x:6 (3 oxygen moles can make 2 Fe₂O₃ moles = x oxygen moles can make 6 Fe₂O₃ moles)

Multiply outside and inside (3*6 , 2*x) and put them on opposing sides of the equation

2*x = 3*6
2x=18
x=9
Therefore 9 moles of oxygen is needed.
The reaction Fe203 occurs when oxygen and iron metal react to each other. The complete balanced formula for the chemical equations will be " 4Fe + 302 -> 2 Fe203. You will need to multiply 6 mols by 3/2 (1.5) and that will give the 9 mols needed to find the number of moles of oxygen that are needed.

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr (at the moment of mixing). The system then re-equilibrates. The appropriate chemical equation is:PCl3(g) + Cl2(g) ---> PCl5(g)

Calculate the new partial pressures after equilibrium is reestablished. [in torr]

PPCl3

PPCl2

PPCl5

Answers

equilibrium is a situation in which economic forces such as supply and demand are balanced and in the absence of external influences the values of economic variables will not change

The equilibrium constant depends on the following:-

  • Pressure
  • Volume

The formula used in the question is as follows:-

k_p = (P_1)/(P_1*P_2), After putting the value, the equilibrium constant is as follows:-

k_p = (217)/(13.2*13.2)

After solving it, the equilibrium constant is 1.245.

The pressure in different systems is as follows:-

Hence, the total pressure is:- P_1 + P_2+ P_3

263 = 13.2 + P_2 + 217

After solving it, the P2 is 32.8torr.

The equilibrium constant in the second case is:-

k_p = (P_1^')/(P_1^' * P_2^')

After putting the value,

1.245= (217+x)/((13.2-x )* (32.8-x))

After solving, the value of x is 6.402torr

Hence, the partial pressure PCL_3,\ CL2, \ PLC_5\ is 6.798, 26.398, and 223.402 respectively.

For more information, refer to the link;-

brainly.com/question/18222206

Answer:

The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=6.798 Torr

The new partial pressures after equilibrium is reestablished Cl_2:

P_2'=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=P_1=13.2 Torr

Partial pressure of the Cl_2=P_2=13.2 Torr

Partial pressure of the PCl_5=P_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=(P_1)/(P_1* P_2)

=(217.0 Torr)/(13.2 Torr* 13.2 Torr)=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=P_1'=13.2 Torr

Partial pressure of the Cl_2=P_2'=?

Partial pressure of the PCl_5=P_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=P_1'+P_2'+P_3'

263.0Torr=13.2 Torr+P_2'+217.0 Torr

P_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=(P_3')/(P_1'* P_2')

1.245=((217.0+x))/((13.2-x)(32.8-x))

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

The new partial pressures after equilibrium is reestablished Cl_2:

P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr
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