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3 years ago

6

Rajah 3 / Diagram 3 Antara industri berikut yang manakah menggunakan mikroorganisma ini? Which of the following industries uses

this microorganism?

2 answers:

tensa zangetsu [6.8K]3 years ago

6 0

Answer:

bacteria is used by industries . Because some bacterias are very friendly

Rudiy273 years ago

5 0

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A 0.239 gram sample of a gas in a 100-ml flask exerts a pressure of 603 mm Hg at 14 Celsius. What is the gas? (Hint: don’t forge

AlekseyPX

Answer:

A) chlorine

Explanation:

To solve this question we can use:

PV = nRT

In order to solve the moles of the gas. With the moles and the mass we can find the molar mass of the gas to have an idea of its identy:

PV = nRT

PV / RT = n

<em>Where P is pressure: 603mmHg * (1atm / 760mmHg) = 0.7934atm</em>

<em>V = 100mL = 0.100L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 14°C + 273.15 = 287.15K</em>

0.7934atm*0.100L / 0.082atmL/molK*287.15K = n

3.37x10⁻³ moles of the gas

In 0.239g. The molar mass is:

0.239g / 3.37x10⁻³ moles = 70.9g/mol

The gas with this molar mass is Chlorine, Cl₂:

<h3>A) chlorine </h3><h3 />

6 0

3 years ago

HELP MEEEEEEE PLSSS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1 As distance increases between a rocket ship and the ground, gravitationa

Sladkaya [172]

Answer:

decreases

Explanation:

Gravirational force is directly proportional to the mass and inversely proportional to the distance.(Newton's law of gravitation)

8 0

3 years ago

A first order reaction has rate constants of 4.6 x 10-2 s-1 and 8.1 x 10-2 s-1 at 0ºC and 20ºC, respectively. What is the value

Airida [17]

Answer:

D. 18,800 J/mol

Explanation:

We need to use the Arrhenius equation to solve for this problem:

$k=Ae^{\frac{-E_a}{RT}$ , where k is the rate constant, A is the frequency factor, $E_a$ is the activation energy, R is the gas constant, and T is the temperature in Kelvins.

We want to find the value of $E_a$ , so let's plug some of the information we have into the equation. The gas constant we can use here is 8.31 J/mol-K.

At 0°C, which is 0 + 273 = 273 Kelvins, the rate constant k is $4.6*10^{-2}$ . So:

$k=Ae^{\frac{-E_a}{RT}$

$4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}$

At 20°C, which is 20 + 273 = 293 Kelvins, the rate constant k is $8.1*10^{-2}$ . So:

$k=Ae^{\frac{-E_a}{RT}$

$8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}$

We now have two equations and two variables to solve for. We just want to find Ea, so let's write the first equation for A in terms of Ea:

$4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}$

$A=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }$

Plug this in for A in the second equation:

$8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}$

$8.1*10^{-2}=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }e^{\frac{-E_a}{8.31*293}$

After some troublesome manipulation, the answer should come down to be approximately:

Ea = 18,800 J/mol

The answer is thus D.

5 0

3 years ago

Calculating the expected pH of the buffer solution: Given that the pKa for Acetic Acid is 4.77, calculate the expected pH of the

ser-zykov [4K]

Answer:

$pH=4.77$

Explanation:

From the question we are told that:

pKa for Acetic Acid $pK_a= 4.77$

Therefore

For Equal Concentration of acetic acid and acetatic ion

$CH_3COOH=CH_3COO^-$

Generally the Henderson's equation for pH value is mathematically given by

$pH=pK_a+log\frac{base}{acid}$

$pH=4.77+log\frac{CH_3COO^-}{CH_3COOH}$

$pH=4.77+log1$

$pH=4.77$

5 0

3 years ago

What is an extensive property of substance?

Helen [10]

Depend on the amount of matter being measured

6 0

4 years ago