What causes the ocean tides to change from high tide to low tide?

Substance A has the following properties.

givi [52]

A curve of temperature vs. time for the entire heating process.

The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.

i ) Calculating the heat and time required to heat the sample to its boiling point:

Boiling point = 85°C

C(liquid) = 2.5 J/g °C

The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,

Therefore, T f = 85°C and T i = - 20°C

Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,

q3 = 25 g × 2.5 J/g °C × [85 - (-20)]°C

= 25 J/°C ×[85+20]°C

= 6562.5 J

The total heat energy required for heating the sample from initial temperature to boiling point is:-

q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J

= 11562.5 J

The Rate of heating = 450 J/min

450. J = 1 min

11562.5 J = ? min

11562.5 J × 1min/450 J = 25.69 min

ii) Calculating the heat and time required to boil the sample:

∆H Vap = 500 J/g

The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined

q4= m × ∆Hvap

= 25 g × 500 J/g

= 12500 J

Thus, total heat required to this phase change is q1 + q2 + q3 + q4 = 500 J + 4500 J +6562.5 J + 12500 J = 24062.5 J

The Rate of heating = 450 J / min

450 J = 1 min

24062.5 J = ? min

24062.5J × 1min / 450 J = 53.47 min

iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C

C gas = 0.5 J / g °C

The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,

Therefore, T f = 100.°C and T i = 85°C

q5 = 25 g × 0.5 J / g °C × [100 - 85] °C

= 25 J / °C ×15 °C

= 187.5 J

The total heat energy required for heating the sample from initial temperature to 100°C is

q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J

=24250 J

The Rate of heating = 450 J / min

450. J = 1 min

24250 J=? min

Thus, heating the sample to 100.°C takes a total of 53.89 min.

iv) Draw a curve of temperature vs. time for the entire heating process:-

Temperature °C Temperature K Heat energy (J) Time (min)

-40 °C 233 0 0

-20 °C 253 500 1.11

Melting -20 °C 253 5000 11.11

85 °C 358 11562.5 25.69

Boiling 85 °C 358 24062.5 53.475

100 °C 373 24250 53.89

Hence, the graph for the result is in the image.

Learn more about temperature here:-brainly.com/question/24746268

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4 0

2 years ago

The necessity of ascorbic acid for its role in _____ is believed to be the reason that people with scurvy have no energy and are

Debora [2.8K]

Answer:

Carnitine Synthesis

Explanation:

Ascorbic acid acts as a cofactor in various hydroxylation and amidation reactions by electron transfer to enzymes that provide reducing equivalents. Accordingly, ascorbic acid is required or facilitates the conversion of certain proline and lysine residues of the procollagen, the oxidation of lysine side chains to proteins, providing hydroxytrimethylsiline for carnitine synthesis. This means that ascorbic acid is essential for the carnitine synthesis.

Carnitine is derived from amino acids and acts as a cofactor for the enzyme acetyltransferase responsible for the transport of fatty acids to mitochondria. Your presence in the body is very important for feeling energetic and strong. The lack of canitine, influenced by the lack of ascorbic acid in individuals suffering from scurvy, causes a lack of energy.

5 0

3 years ago

Calculate the concentration of CO2 in water at 25 degrees Celsius when the pressure of CO2 over the solution is 4.5 atm. At 25 d

OLEGan [10]

Answer : The concentration of $CO_2$ is, 0.12 M

Explanation :

Using Henry's law :

$C_{CO_2}=k_H\times p_{CO_2}$

where,

$C_{CO_2}$ = concentration of $CO_2$ = ?

$p_{CO_2}$ = partial pressure of $CO_2$ = 4.5 atm

$k_H$ = Henry's law constant = $3.1\times 10^{-2}mol/L.atm$

Now put all the given values in the above formula, we get:

$C_{CO_2}=3.1\times 10^{-2}mol/L.atm\times (4.5atm)$

$C_{CO_2}=0.1395M\approx 0.12M$

Thus, the concentration of $CO_2$ is, 0.12 M

5 0

3 years ago

to form brass, copper and zinc combine but do not chemically bond. brass is a(n) ________. a atom b element c compound d

umka21 [38]

Answer: Mixture

Explanation:To form brass, copper and zinc combine but do not chemically bond. Brass is a(n)

3 0

3 years ago

If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th

qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

$K_c=4.74\times 10^4$

Concentration at equilibrium:

$[COBr_2]=1.58\times 10^{-6}M$

$[Co]=2.78\times 10^{-3}M$

$[Br_2]=2.51\times 10^{-5}M$

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

The expression for reaction quotient will be :

$Q=\frac{[CO][Br_2]}{[COBr_2]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}$

The given equilibrium constant value is, $K_c=4.74\times 10^4$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K_c$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the $Q$ that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0

3 years ago