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3 years ago

Describe each of the 3 main domains used to classify living things.

Chemistry

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

Answer:

Organisms can be classified into one of three domains based on differences in the sequences of nucleotides in the cell's ribosomal RNAs (rRNA), the cell's membrane lipid structure, and its sensitivity to antibiotics. 3. The three domains are the Archaea, the Bacteria, and the Eukarya.

Explanation:

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Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

3 years ago

A sample of Hydrogen gas has a volume of 25.5 liters at a pressure of 15.0 kPa. If the temperature is kept constant and the pres

patriot [66]

Answer:

<h2>7.65 litres </h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we're finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

15 kPa = 15,000 Pa

50 kPa = 50,000 Pa

We have

$V_2 = \frac{15000 \times 25.5}{50000} = \frac{382500}{50000} \\ = 7.65$

We have the final answer as

<h3>7.65 litres </h3>

Hope this helps you

5 0

3 years ago

What is the volume of NH3 produced in the following reaction when 3.0L of N2 reacts with 4.0L of H2?

trapecia [35]

Hey there !

Given the reaction:

N2 + 3 H2 = 2 NH3

At constant pressure and temperature ,volume is proporcional to moles:

Theoretical moles of N2 and H2 => 1:3

Theoretical volume of N2 and H2 => 1:3

Experimental volume of N2 and H2 => 3.0 L : 4.0 L

0.75 : 1 = 2.25 : 3

Since N2 is in excess reactant , H2 is the limiting reactant

Therefore:

volume of NH3 is 2/3 * Volume of H2

= 2/3 * 4.0 = 2.66 L

Hope that helps!

4 0

4 years ago

Read 2 more answers

If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

hodyreva [135]

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $Al_2O_3$ = $\frac{16.9\times 1000g}{102g/mol}=165.7moles$

moles of $NaOH$ = $\frac{57.4\times 1000g}{40g/mol}=1435moles$

moles of $HF$ = $\frac{57.4\times 1000g}{20g/mol}=2870moles$

As 1 mole of $Al_2O_3$ reacts with 6 moles of $NaOH$

166 moles of $Al_2O_3$ reacts with = $\frac{6}{1}\times 166=996$ moles of $NaOH$

As 1 mole of $Al_2O_3$ reacts with 12 moles of $HF$

166 moles of $Al_2O_3$ reacts with = $\frac{12}{1}\times 166=1992$ moles of $HF$

Thus $Al_2O_3$ is the limiting reagent.

As 1 mole of $Al_2O_3$ produces = 2 moles of cryolite

166 moles of $Al_2O_3$ reacts with = $\frac{2}{1}\times 166=332$ moles of cryolite

Mass of cryolite $(Na_3AlF_6)$ = $moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg$

Thus 69.72 kg of cryolite will be produced.

8 0

3 years ago

6547.9 feet to meters<br> Using dimensional analysis

ELEN [110]

Answer:

92.964

Explanation:

8 0

3 years ago