Doctors are investigating the case of a patient with defective proteins. Which organelle of the cell is most likely the cause?

lysosome
cell membrane
cytoplasm
ribosome

306.178 liters is the volume of O2 at STP, required for the complete combustion of 125g octane (\(C_{8}H_{18}\)) to \(CO_{2}\) and H20

To calculate the volume of \(O_{2}\) required for the complete combustion of octane (\(C_{8}H_{18}\)) to \(CO_{2}\) and \(H_{2}O\) at STP (Standard Temperature and Pressure), we need to consider the stoichiometry of the balanced chemical equation.

The balanced equation for the combustion of octane is:

\(C_{8}H_{18}\) + 12.5\(O_{2}\) -> 8\(CO_{2}\) + 9\(H_{2}O\)

From the equation, we can see that 1 mole of octane requires 12.5 moles of \(O_{2}\) to completely combust. The molar mass of octane (\(C_{3}H_{18}\)) is approximately 114.22 g/mol.

To calculate the moles of octane, we divide the given mass by the molar mass:

Moles of octane = 125 g / 114.22 g/mol ≈ 1.093 mol

Since the molar ratio between octane and \(O_{2}\) is 1:12.5, the moles of \(O_{2}\)required can be calculated as:

Moles of \(O_{2}\) = 1.093 mol * 12.5 ≈ 13.663 mol

Now, we can use the ideal gas law, PV = nRT, to calculate the volume of \(O_{2}\) at STP. At STP, the temperature is 273 K, and the pressure is 1 atm.

Using the molar volume of an ideal gas at STP (22.4 L/mol), the volume of \(O_{2}\) required is:

Volume of \(O_{2}\) = Moles of \(O_{2}\) * Molar volume = 13.663 mol * 22.4 L/mol ≈ 306.178 L

Therefore, approximately 306.178 liters of \(O_{2}\) at STP would be required for the complete combustion of 125 grams of octane (\(C_{8}H_{18}\)) to \(CO_{2}\) and \(H_{2}O\)

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Answer:

Answer:

See below ~

Explanation:

Finding the rule of the function :

Inverse rule :

Table :

\(\left[\begin{array}{ccc}x&y\\-2&-\frac{1}{2} \\-3&-1\\-4&0\\-5&1\end{array}\right]\)

Answer:

1.96mL

Explanation:

Density = mass/volume, and rearranged to solve for volume, volume = mass/density.

So:

volume = 5.30g/2.70g/mL = 1.96mL (assuming your unit was g/mL for density)

OSHA has modified its Hazard Communication Standard's labeling rules for dangerous chemicals.  

the label may declare, "Do not inhale vapors or spray. Labels for hazardous chemicals must contain: Name, Address, and Telephone Number; Product Identifier; Signal Word; Hazard Statement(s); Precautionary Statement(s); and Pictogram(s).If you feel ill, get medical attention.

A product identifier, a signal phrase, and a pictogram to depict the hazards are required on chemical labels.

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Answer:

F,E,D,C, B, A, G, H, I

Explanation:

I think. It goes bottom to top. the fault probably cause g to happen and maybe h but mattering on what caused what I am not positive

Answer:

6.97 atm is the osmotic pressure.

The plants are disinfectant by adding chlorine. Hence, option A is the correct answer.

Chlorine is found to have disinfectant abilities when it is used on plants, so plants are disinfectant by adding chlorine. It helps to keep the plants safe from viruses and fungus. It operates by destroying the enzymes and genetic material of microbes by breaking the cell membrane.  

However, it is critical to apply chlorine at the optimum quantity and to thoroughly rinse it off to avoid any negative effects on the quality and safety of food products.

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This element is found in group 3A, period 3

The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)

Electron configuration of element X : 2.8.3 , so :

K shell = 2 ⇒1s²

L shell = 8⇒2s²2p⁶

M shell = 3⇒ 3s²3p¹

Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids

The outer shell 3s²3p¹ : located in group 3A and period 3

group⇒valence electron ⇒3

period⇒the greatest value of the quantum number n⇒3

The final concentration of the reactant of a first order reaction can be determined from the rate constant equation. The concentration of ammonium carbonate after 39 s will be 0.003 M.

Rate constant of a reaction is the rate of reaction when one molar concentration of the reactant is involved in the reaction. The expression for the rate constant k for first order reaction is :

k = 1/t ln (C0/Ct)

Where C0 be the initial concentration and Ct be the concentration after t seconds.

Given that C0 of ammonium nitrate = 0.957 M

rate constant = 0.196 /s

t = 39 s.

The concentration after 39 seconds is calculated as follows:

0.196 /s = 1/39s ln (0.957 M / Ct)

Ct = 0.957 / (ln⁻¹ (0.196 × 39))

    = 0.003 M.

Therefore, the concentration of ammonium carbonate after 39 seconds will be 0.003 M.

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The value of the rate constants at various temperatures can be measured from the plot of ln K versus 1/T. The activation energy of the reaction obtained from the plot is

The minimum excess energy that the reactants must acquire so as to have energy equal to the threshold energy for the reaction is defined as the activation energy.

In the plot of ln K versus 1/T the slope obtained is [ -Eₐ / 2.303 R ] and the intercept on the y -axis is equal to log A.

Here the slope in the plot is given as − 5780 K. Then the activation energy is calculated as:

Slope = -Eₐ / R

-5780 K = -Eₐ / 8.314 J/ mol.K

Eₐ =  8.314 J/ mol.K × 5780 K = 48054.9 J/ mol = 48.054 kJ.

Thus the activation energy of the reaction is 48054.9 J/ mol or 48.054 kJ.

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Answer:

\(4.92x10^{23}molecules H_2O\)

Explanation:

Hello,

In this case, since the chromium(III) oxalate trihydrate is Cr₂(C₂O₄)₃ ·3H₂O whose molar mass is 422 g/mol and one mole of chromium(III) oxalate trihydrate contains three moles of water, by also considering that one mole of substance contains the Avogadro´s number in particles, the number of water molecules turns out:

\(=115gCr_2(C_2O_4)_3\ 3H_2O*\frac{1molCr_2(C_2O_4)_3\ 3H_2O}{422gCr_2(C_2O_4)_3\ 3H_2O}*\frac{3molH_2O}{1molCr_2(C_2O_4)_3\ 3H_2O} *\frac{6.022x10^{23}molecules H_2O}{1molH_2O} \\\\=4.92x10^{23}molecules H_2O\)

Best regards.

Based on the information provided, we have a reaction between hydrogen iodide (HI) gas and hydrogen gas (H₂) to form iodine gas (I₂). The equilibrium is represented by the equation:

2HI(g) = H₂(g) + I₂(g)

The concentration values given in the table correspond to the concentrations of H₂ and HI at different times.

a) From t=0 s to t=5 s: Without the specific graph mentioned in Figure 7.10, it is difficult to provide a precise explanation of the physical situation in the container during this time period. However, based on the equilibrium reaction given, we can make some general observations. At the start (t=0 s), the concentrations of H₂ and HI may be high. As time progresses, the reaction proceeds, and the concentrations of H₂ and HI may decrease while the concentration of I₂ increases. The specific behavior will depend on the rate of the forward and reverse reactions.

b) External factor altered at t=5 s: To bring about a change in the shape of the graph at t=5 s, some external factor must have been altered. The most likely factor is the total pressure within the container. Since the reaction involves gases, changes in pressure can affect the equilibrium position. However, according to the information given, a change in pressure will not affect equilibrium in this case since the number of moles of gas is the same on both sides of the equation. Therefore, if the shape of the graph changes at t=5 s, some other external factor, such as temperature or the addition of a catalyst, must have been altered.

c) Calculation of K at t=3 s: The equilibrium constant (K) can be calculated at any given time using the concentrations of the reactants and products. However, the concentrations of H₂ and HI at t=3 s are not provided in the information given. Without the necessary data, it is not possible to calculate K at t=3 s.

Lastly, the statement "1 dm³ COCI, decomposes" seems incomplete. If you provide additional information or clarify the question, I'll be happy to assist you further.

Answer:

When someone spends a long period in a closed well, their ability to breathe becomes more difficult due to the high concentration of carbon dioxide that has accumulated, which could result in death.

can I please get the five points :)

try and find a better answer or ask ur teacher please

Time increases as the volume of hydrochloric acid remains the same. This is because the concentration of hydrochloric acid would remain constant, affecting the reaction rate. Option B

Based on factors that affect the rates of chemical reactions, the expected trend in the table can be analyzed.

In this experiment, calcium carbonate reacts with hydrochloric acid to produce carbon dioxide gas. The reaction rate is influenced by factors such as concentration, surface area, temperature, and presence of a catalyst. Let's examine the options provided:

A. Time increases as the volume of hydrochloric acid decreases:

This option suggests that as the volume of hydrochloric acid decreases, the time taken for carbon dioxide to form increases. However, in general, a decrease in the volume of hydrochloric acid would result in an increase in its concentration.

B. Time increases as the volume of hydrochloric acid remains the same:

If the volume of hydrochloric acid remains the same, it means the concentration of hydrochloric acid is constant. In this case, the reaction rate is not expected to change significantly, and the time taken for carbon dioxide to form should remain relatively constant. Therefore, this option is plausible.

C. Time decreases as the mass of calcium carbonate decreases:

If the mass of calcium carbonate decreases, the surface area of the reactant particles decreases. A smaller surface area would result in less contact between reactant particles, leading to a slower reaction rate. Thus, the time taken for carbon dioxide to form would be expected to increase, making this option unlikely.

D. Time decreases as the mass of calcium carbonate remains the same:

If the mass of calcium carbonate remains the same, the surface area of the reactant particles also remains the same. Therefore, the reaction rate should remain relatively constant, and the time taken for carbon dioxide to form would not show a significant change. Thus, this option is also plausible.

Option B is correct.

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Answer:

17202.6 years

Explanation:

Activity of the living sample (Ao) = 160 counts per minute

Activity of the wood sample (A) = 20 counts per minute

Half life of carbon-14 = 5730 years

t= age of the artifact

From;

0.693/t1/2= 2.303/t log Ao/A

Then;

0.693/ 5730= 2.303/t log Ao/A

Substituting values;

0.693/5730= 2.303/t log (160/20)

Then we obtain;

1.209×10^-4 = 2.0798/t

t= 2.0798/1.209×10^-4

Thus;

t= 17202.6 years

Therefore the artifact is 17202.6 years old.

The new volume of the gas sample is approximately 3.6 L.

According to the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in Kelvin. If pressure is constant, we can use the formula V1/T1 = V2/T2 to find the new volume of the gas sample.

V1/T1 = V2/T2

V2 = (V1 x T2)/T1

V2 = (3 L x 300 K) / 250 K

V2 = 3.6 L

Therefore, the new volume of the gas sample is approximately 3.6 L. As the temperature of the gas sample increased from 250 K to 300 K, the volume increased proportionally since pressure was held constant.

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Metamorphic rock formation involves the transformation of existing rocks under heat and pressure. They form without melting and can appear foliated or non-foliated.

Metamorphic rocks are formed from the transformation of existing rock types in a process called metamorphism, which means 'change in form'. The appropriate phrases that describe metamorphic rock formation are: 'form from existing rocks', 'form without melting', 'appear foliated or non-foliated', and 'require heat and pressure to form'. These rocks are subject to conditions of heat and pressure that cause them to change physically and/or chemically, resulting in a new type of rock. They can either be foliated (layered) or non-foliated. Importantly, metamorphic rock formation does not include a liquid state, meaning they do not 'form from liquid rock' or 'form from deposition'.

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Answer:

A: 10 moles of water are required.

B: 27 moles of C₂H₄  are used.

Explanation:

A:

Given data:

Moles of water used = ?

Moles of C₂H₅OH formed = 10 mol

Solution:

Chemical equation:

C₂H₄ + H₂O     →       C₂H₅OH

Now we will compare the moles of water with C₂H₅OH from balance chemical equation:

                    C₂H₅OH            :           H₂O  

                         1                   :             1

                       10                  :         10

10 moles of water are required.

B)

Given data:

Moles of C₂H₄ used = ?

Moles of C₂H₅OH formed = 27 mol

Solution:

Chemical equation:

C₂H₄ + H₂O     →       C₂H₅OH

Now we will compare the moles of C₂H₄  with C₂H₅OH from balance chemical equation:

                    C₂H₅OH         :           C₂H₄  

                         1                 :             1

                         27               :            27

27 moles of C₂H₄  are used.

Answer:

56.96 grams

Explanation:

To find the mass of 1.78 moles of O2, we need to use the molar mass of O2, which is the mass of one mole of O2.

The chemical formula for O2 is O-O or simply O2. The molar mass of O2 is the sum of the atomic masses of two oxygen atoms, which can be found on the periodic table.

The atomic mass of oxygen (O) is approximately 16.00 g/mol. So the molar mass of O2 is:

Molar mass of O2 = 2 x atomic mass of O

= 2 x 16.00 g/mol

= 32.00 g/mol

Therefore, the mass of 1.78 moles of O2 is:

Mass = number of moles × molar mass

= 1.78 mol × 32.00 g/mol

= 56.96 g

So the mass of 1.78 moles of O2 is 56.96 grams.

Answer:

a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change.

b. ΔS : 3. Change in entropy.

c. Keq : 6. Equilibrium constant.

d. ΔH : 7. Change in enthalpy.

e. ΔG° : 9. Standard free energy change.

f. J : 8. Joule.

Explanation:

Hello,

In this case, we can match the symbol with the proper definition as shown below:

a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change and it uses G since it is better referred to the Gibbs free energy.

b. ΔS : 3. Change in entropy.

c. Keq : 6. Equilibrium constant.

d. ΔH : 7. Change in enthalpy.

e. ΔG° : 9. Standard free energy change.

f. J : 8. Joule.

Best regards.

By using Avogadro's Law we have:

Where V is volume and n is moles, By substituting what have to find the unknown V2 gives:

11.0 L/ 7.70 moles = V2/ 2.97 moles

V2 = 11.0 L x 2.97 moles / 7.7 moles

V2 = 4.24 L. This is the volume of gas that would have escaped

New volume of the container is:

V= 11.0 L- 4.24 L

V= 6.76 L

Answer is 6.76 L

Answer:

\(\boxed {\boxed {\sf 0.0769 \ M}}\)

Explanation:

We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:

\(M_AV_A= M_B V_B\)

In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.

\(M_A * 25.0 \ mL = M_BV_B\)

The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.

\(M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL\)

We are solving for the molarity of the acid and we must isolate the variable \(M_A\). It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

\(\frac {M_A * 25.0 \ mL }{25.0 \ mL}= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}\)

\(M_A= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}\)

The units of milliliters cancel.

\(M_A= \frac{0.108 \ M * 17.80 }{25.0 }\)

\(M_A= \frac{1.9224}{25.0 } \ M\)

\(M_A= 0.076896 \ M\)

The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.

\(M_A \approx 0.0769 \ M\)

The molarity of the hydrochloric acid is 0.0769 Molar.

The final temperature of the metal is 1.88°C, while the final temperature of water is 50°C. The final temperature of both is 25.94°C (average of 1.88°C and 50°C).

To answer the question, it is necessary to use the formula: Q = m * c * ΔT. The formula is used to calculate the amount of heat gained or lost by a substance. Here, the heat lost by the metal will be equal to the heat gained by water. Let's use the following terms in our answer:
- ΔT: change in temperature
- m: mass of substance
- c: specific heat capacity

Initially, the metal has a higher temperature, while the water has a lower temperature. Heat energy will flow from the metal to the water until both substances reach a common temperature.

The rate of heat transfer will be directly proportional to the difference in temperature between the two substances. To observe the increase in temperature every 60 seconds for 300 seconds, we will make a table with time intervals and corresponding temperature readings.

Finally, we can use the formula to find the final temperature of the metal and water. Let's assume that the mass of water and metal are equal, and both are 100 grams. The specific heat capacity of water is 4.18 J/g °C, while the specific heat capacity of the metal is 0.91 J/g °C.

Heat lost by metal = Heat gained by waterm * c * ΔT(metal) = m * c * ΔT(water)100 * 0.91 * (50 - T) = 100 * 4.18 * (T - 20) 4550 - 91T = 418T - 8364.37T = 83.64T = 1.88°C

Therefore, the final temperature of the metal is 1.88°C, while the final temperature of water is 50°C. The final temperature of both is 25.94°C (average of 1.88°C and 50°C).

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Heat would be required  : 1,670 J

Given

mass of H₂O=5 g

Required

Heat to melt

Solution

The heat to change the phase can be formulated :

Q = m.Lf (melting/freezing)

Lf=latent heat of fusion

The heat of fusion for water at 0 °C : 334 J/g

Input given values in formula :

\(\tt Q=5\times 334=\boxed{\bold{1,670~J}}\)

The ΔG for the reaction shown is -11.76 kJ/mol.

The equilibrium constant is a value that shows the extent to which reactants are converted to products in a reaction. We have to first obtain the equilibrium constant as follows;

K = (0.750)^2/(0.160) (0.100) = 35

Now;

Using the formula;

ΔG = -RTlnK

ΔG = -(8.314 × 398 × ln 35)

ΔG = -11.76 kJ/mol

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