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2 years ago

2. List the following from least complex to most complex.

Chemistry

2 answers:

nekit [7.7K]2 years ago

7 0

Answer:

Bacterial cell, cell with a nucleus, multicellular organism

poizon [28]2 years ago

7 0

It’s bacterial cell then cell with nucleus the multicellular organism

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Based on the current atomic theory most chemical reaction take place as a result of the interactions of which sub- atomic partic

icang [17]

Electrons, everything is pretty much based around the likeliness of electrons to be swapped or shared between atoms

6 0

2 years ago

A laboratory technician combined sodium hydroxide with excess iron(II) nitrate. A reaction took place according to this chemical

Whitepunk [10]

The initial mass of sodium hydroxide is 3.3 g (answer C)

<u><em>calculation</em></u>

Step 1 : find the moles of iron (ii) hydroxide ( Fe(OH)₂

moles = mass÷ molar mass

from periodic table the molar mass of Fe(OH)₂ = 56 + [16 +1]2 = 90 g/mol

moles is therefore = 3.70 g÷ 90 g/mol = 0.041 moles

Step 2: use the mole ratio to calculate the moles of sodium hydroxide (NaOH)

from given equation NaOH : Fe(OH)₂ is 2 :1

therefore the moles of NaOH = 0.041 x 2 = 0.082 moles

Step 3: find mass of NaOH

mass = moles x molar mass

from the periodic table the molar mass of NaOH = 23 +16 +1 = 40 g/mol

mass = 0.082 moles x 40 g/mol = 3.3 g ( answer C)

8 0

3 years ago

If 75 grams of oxygen react, how many grams of aluminum are required?

german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

O₂ : Al

3 : 4

2.34 : 4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

5 0

2 years ago

How are sea slugs able to make their own food like green algae?

Gala2k [10]

C. They are genetically capable of making chlorophyll

3 0

2 years ago

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

2 years ago