1) The Specific Heat Capacity at 25°C of Gold is 0.129 / (g°C)

a) What is the molar Specific Heat Capacity of Gold in J/(mol °C)​

Answer:

1. Distance  = 5 cm

Displacement = 1 cm

2. Distance = 7 cm

Displacement = 1 cm

3. Distance = 5 cm

Displacement = -1 cm

4. Distance  = 6 cm

Displacement = -2 cm

5. Distance  = 4 cm

Displacement = 0 cm

Explanation:

The distance an object travels is a measure of the total ground cobpvered bybthe object without regard to the direction of motion. It is a scalar quantity as it only takes into account magnitude but not the direction of a quantity.

Displacement however, is a measure of how far away or apart an object is from its starting position.  It a vector quantity as it takes into account both the magnitude and direction in its measurement.

From the attachment provided, considering forward direction as positive and reverse as negative :

1. When a object moves from point 1 to point 4 and reverses to point 2.:

Distance  = from 1 to 4 = 3 cm; from 4 back to 2 = 2 cm;  3 + 2 = 5 cm

Displacement = from 1 to 4 = +3 cm; from 4 back to 2 = -2 cm;  3 - 2 = 1 cm

2. When an object moves from point 1 to 5 and reverses to 2:

Distance = from 1 to 5 = 4cm; from 5 back to 2 = 3 cm; 4 + 3 = 7 cm

Displacement = from 1 to 5 = +4 cm; from 5 back to 2 = -4 cm;  4 - 3 = 1 cm

3. When an object moves from point 1 to 3 and then reverses to 0.

Distance  = from 1 to 3 = 2 cm; from 3 back to 0 = 3 cm;  2 + 3 = 5 cm

Displacement = from 1 to 3 = +2 cm; from 3 back to 0 = -3 cm;  2 - 3 = -1 cm

4. When an object moves from point 3 to 5 and then reverses to 1:

Distance  = from 3 to 5 = 2 cm; from 5 back to 1 = 4 cm;  2 + 4 = 6 cm

Displacement = from 3 to 5 = +2 cm; from 5 back to 1 = -4 cm;  2 - 4 = -2 cm

5. When an object moves from point 2 to 4 and then reverses to 3.

Distance  = from 2 to 4 = 2 cm; from 4 back to 2 = 2 cm;  2 + 2 = 4 cm

Displacement = from 2 to 4 = +2 cm; from 4 back to 2 = -2 cm;  2 - 2 = 0 cm

Answer:

B

Explanation:The poisonous substances, present in the environment can easily get into the trophic level as living organism depends on each other and environment for food and nutrition. These poisonous substances may not be broken down in the body or excreted easily, efficiently and quickly. Instead, they accumulate in the tissues, and as the living organism eats more, the concentration of these substances increases and they pass from one trophic level to the next. The tertiary consumer being at the top of trophic levels receives the maximum pollutant. This phenomenon is known as biological magnification.

So, the correct answer is option B.

Answer:

1.77 * 10^-3

Explanation:

From the titration formula;

Let

CA = concentration of acid

CB = concentration of base

VA = volume of acid

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

The equation of the reaction is;

M(OH)2(aq) + 2HCl(aq) -------> MCl2(aq) + 2H2O(l)

So;

CAVA/CBVB= NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 0.173 * 25.10 * 1/28.5 * 2

CB= 4.3423/57

CB= 0.0762 M

This implies that the solubility of M(OH)2 = 0.0762 M

M(OH)2(s) ----> M^+(aq) + 2OH^-(aq)

So

Ksp = x * (2x)^2

Ksp = 4x^3

x = 0.0762

Ksp= 4(0.0762)^3

Ksp = 1.77 * 10^-3

Answer:

36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 80 °C

T(K) = 80 + 273

T(K) = 353 K

Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

Number of mole (n) = 1.27 moles

Temperature (T) = 353 K

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

1 × V = 1.27 × 0.0821 × 353

V = 36.8 L

Thus, the volume occupied by the helium gas is 36.8 L

Answer: Primary consumers make up the second trophic level. They are also called herbivores. They eat primary producers—plants or algae—and nothing else. For example, a grasshopper living in the Everglades is a primary consumer

Answer:

primary consumer make up the second trophic level .they are also herbivores they eat primary consumer plants or alger and nothing else .for example a grasshopper living in the everglades is a primary consumer

Answer:

CuSO4 cell will have the greatest amount of deposit among all three. The deposit will occur at the cathode

Explanation:

The valence of the elements in this case is as follows -

Cu - 2e-

Sn - 4e-

Cr - 3e-

CuSO4 cell will have the greatest amount of deposit among all three

The atoms of copper metal will deposit at the cathode. At the cathode, the least number of moles of electrons needed .

Hence, more amount of copper  can be extracted out by the electrolyte

Helium, bromine, agron, neon,flourine

Answer:

At D: Soil erosion us the process taking place.

Soil erosion is the washing away of the Earth's or soil's top layer by eroding agents.

Answer:

Ka=[H+][A−]/[HA]

which solution has the lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq) :

Higher value of Ka indicates greater ionization thus more protons in solution, higher H+ concentration in solution means lower pH value.

Since HF have higher Ka have, thus HF solution will have lower pH.

The measurement is when the solution has the lower pH so HF should have higher ka.

In the given solution, it contains the  lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq).

Here higher value of Ka represents the high ionization so there should be more protons in the solutions. and, the higher h+ concentration have the lower pH value.

So in the case when HF has higher Ka, due to this, HF solution will have lower pH.

Learn more about solution here: https://brainly.com/question/14619461

Solution :

The balanced chemical equation are

The Catabolism of the glucose takes place in five stages :

1. Glucose + [tex]$ATP$[/tex] → glucose - [tex]$6$[/tex] - phosphate [tex]$+ ADP$[/tex]  ,  [tex]$\Delta G^0=-16.7 \ kJ/mol$[/tex]

2. Glucose - [tex]$6$[/tex] - phosphate → Fructose - [tex]$6$[/tex] - phosphate ,  [tex]$\Delta G^0=1.7 \ kJ/mol$[/tex]

3. Glucose - [tex]$6$[/tex] - phosphate + [tex]$ATP$[/tex] → [tex]$ADP$[/tex] + Fructose - [tex]$1,6- \text{biophosphate}$[/tex],  [tex]$\Delta G^0=-14.2 \ kJ/mol$[/tex]

4. Fructose - [tex]$1,6- \text{biophosphate}$[/tex]  → dihydroxyacetonephosphate + glyceraldehyde  [tex]$-3 - $[/tex] phosphate,   [tex]$\Delta G^0=23.8 \ kJ/mol$[/tex]

5. Dihydroxyacetonephosphate → glyceraldehyde  [tex]$-3 - $[/tex] phosphate , [tex]$\Delta G^0=7.5 \ kJ/mol$[/tex]

Therefore, the overall net equations

Glucose + [tex]$2ATP \rightarrow$[/tex] glyceraldehyde  [tex]$-3 - $[/tex] phosphate + [tex]$2ADP$[/tex] [tex]$\Delta G^0=2.1 \ kJ/mol$[/tex]

Answer:

https://sensorex.com/ph-calculator/

Explanation:

go to this link and then scroll down a little bit and then you'll find the answer

Answer:

[tex]\Delta H_{comb}=2043.85kJ/mol[/tex]

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

[tex]\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}[/tex]

Thus, given the values of the enthalpies of formation on the attached file, we obtain:[tex]-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol[/tex]

Best regards!

Answer:

Explanation:

Heat is a reactant.

If the equation requires heat to occur, the reaction is endothermic

The equation would have this model

A + B + heat =====> C + D

If the equation gives away heat, then it is exothermic

The equation would look like this. (I'll use a common reaction)

CH4 + 2O2 - heat ===> CO2 + 2H2O

The heat is minus when it has to be given away.

Answer:

Temperature usually increases when water condenses. What behavior of water is most directly responsible for this phenomenon? The release of heat by the formation of hydrogen bonds. Hydrogen bonds stabilize and keep the of ice farther apart than the water molecules of liquid water.

Answer:

B is correct :)

Explanation:

Trust me I just took the test

Answer:

C is the correct answer.

Explanation:

We can eliminate some answers immediately:

A is incorrect because nitrogen and hydrogen do not change their identity, they just combine into a new molecule that contains them. Notice how there are 2 nitrogen atoms and 6 hydrogen atoms on both sides of the equation.

B is incorrect because the first law of thermodynamics tells us that matter and energy will always be conserved, so the mass of the products must be equal to the mass of the reactants, not less than the mass of the reactants.

C is correct because balanced chemical equations help chemists predict how much product will form from certain amounts of reactants.

D is incorrect because this equation is showing the exact opposite. Molecular hydrogen and nitrogen can combine stoichiometrically to form ammonia.

Answer:

See explanation

Explanation:

According to Bronsted-Lowry, an acid is a proton donor while a base is a proton acceptor.

Hence, if we consider the reaction above, we will notice that for each base there is a conjugate acid and for each acid there is a conjugate base.

For the acid HNO3, its conjugate base is NO3^- while for the acid H3O^+, its conjugate base is H2O.

Answer:

See Explanation

Explanation:

An ionic bond occurs due to electrostatic attraction between a positively charged ion and a negatively charged ion.

A metal and a ligand are bound by a coordinate covalent bond or a dative bond. This bond occurs due to donation of electron pairs from ligands to available orbitals on metals.

The formation of coordinate bonds is evident when neutral molecules or negative ions with non bonding electrons donate same to empty metal orbitals. This is sometimes shown by an arrow pointing from the ligands to the metal center.

For instance; tetraammine copper II ion is formed when four ammonia molecules donate a lone pair each to available vacant orbitals of the copper metal center to form [Cu(NH3)4]^2+.

Answer:

The answer is high entropy.

Explanation:

Answer:

1 mole of KClO_3 contains 3 moles of O.

10 moles of O_2 contains 20 moles of O.

You need 20/3 moles of KClO_3 to produce 10 moles of O_2

Explanation:

sana makatulong

Answer:

38.4%

Explanation:

We'll begin by calculating the molar mass of Al(ClO₃)₃. This can be obtained as follow:

Molar mass of Al(ClO₃)₃

= 27 + 3[35.5 + (16×3)]

= 27 + 3[35.5 + 48]

= 27 + 3[83.5]

= 27 + 250.5

= 277.5 g/mol

Next, we shall determine the mass of Cl in Al(ClO₃)₃. This can be obtained as follow:

Mass of Cl in Al(ClO₃)₃ = 3 × Cl

= 3 × 35.5

= 106.5 g

Finally, we shall determine the percentage of Cl in Al(ClO₃)₃. This can be obtained as follow:

Mass of Cl in Al(ClO₃)₃ = 106.5 g

Mass of Al(ClO₃)₃ = 277.5 g

Percentage of Cl =?

Percentage of Cl = mass of Cl / mass of Al(ClO₃)₃ × 100

Percentage of Cl = 106.5 / 277.5 × 100

Percentage of Cl = 38.4%

Answer:

habitat/shelter

Explanation:

Answer:

Look below

Explanation:

Nitrogen: N

Phosphorus: P

Bromine: Br

Zinc: Zn

Potassium: K

Magnesium: Mg

Hope this helps(would be very grateful if you mark this answer branliest bc i need for lvl up)

Answer:

12.8

Explanation:

14 = pOH + pH

pH = 14 - pOH

pH = 14 - 1.2

pH = 12.8

Answer:

1) increase

2) increase

Explanation:

Entropy is the degree of disorderliness or randomness of a system. It is the measure of the unavailable energy in a system.

Entropy increases with increase in the number of particles. If the number of particles in a system increases from left to right, the entropy of the system increases accordingly.

In reaction 1, the number of particles from left to right increased from two to three hence the entropy was increased.

In reaction 2, the number of particles from left to right increased from three to five hence the entropy was increased.

Answer:

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]

Explanation:

From the question we are told that

Light wavelength  [tex]\lambda_l=655nm[/tex]

Light wavelength atom fall [tex]x_L=5th-2nd[/tex]

Photon wavelength  [tex]\lambda_p=435nm[/tex]

Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]

Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by

[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]

Therefore for initial drop of 5th to 2nd

[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]

Therefore for initial drop of 6th to 2nd

[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

Generally we subtract (5th to 2nd) from (6th to 2nd)

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]

Therefore for 6th to 5th stage is mathematically given by

[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]

[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]