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3 years ago

The System that Linnaeus classifies organisms by? A.genus and order B.order and species C.genus and species D.Kingdom and class

Chemistry

2 answers:

My name is Ann [436]3 years ago

8 0

C. genus and species

I hope this helped!

Eva8 [605]3 years ago

4 0

Hello, Umarhaley!

The answer to your amazing question is genus and species.

Have a good day! -Wajiha

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What is SI unit of e.m.f and p.d.

xenn [34]

Answer:

Electromotive force or e.m.f is defined as the battery's energy per Coulomb of charge passing through it. like other measures of energy per charge emf has SI unit of volts , equivalent to joules per coulomb.

8 0

3 years ago

Concentrated hydrochloric acid, HCl, comes with an approximate molar concentration of 12.1 M. If you are instructed to prepare 3

Semmy [17]

Answer:

28.20 mL of the stock solution.

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 12.1 M

Volume of diluted solution (V2) = 350.0 mL

Molarity of diluted solution (M2) = 0.975 M

Volume of stock solution needed (V1) =..?

The volume of stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12.1 x V1 = 0.975 x 350

Divide both side by 12.1

V1 = (0.975 x 350)/12.1

V1 = 28.20 mL.

Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.

6 0

3 years ago

30 POINTS! ----- How many moles of oxygen gas are needed to completely react with 145 grams of aluminum? Report your answer with

gladu [14]

First a balanced reaction equation must be established:
$4Al _{(s)} + 3 O_{2} _{(g)}$ → $2 Al_{2} O_{3}$

Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols

Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]

∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)

Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.

4 0

4 years ago

What is [h3o ] in a solution of 0.075 m hno2 and 0.030 m nano2?

Tanya [424]

Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use Henderson–Hasselbalch equation:
pH = pKa + log(c(NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
[H₃O⁺] = 10∧(-4,02).
[H₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.

8 0

3 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.