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2 years ago

A __________ is the basic unit of structure and function in living organisms.

Chemistry

2 answers:

Llana [10]2 years ago

6 0

The correct answer is Cell

Soloha48 [4]2 years ago

5 0

The answer is cell, a cell is the basic structure function in a living organisms

Hope this helps you out.

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What are elements that bond with actinium

sineoko [7]

Fluorine, chlorine, iodine (the halogens), oxygen, sulphur, hydrogen

5 0

3 years ago

Consider the chemical equation for the production of water: 2 H2+O2→2 H2O. If 100 grams of oxygen gas are used, what would the p

Cerrena [4.2K]

Answer:

The percentage yield of water is 66.67%.

Explanation:

$2H_2+O_2\rightarrow 2H_2O$

Mass of oxygen gas = 100 g

Moles of oxygen gas = $\frac{100 g}{32 g/mol}=3.125 mol$

According to reaction, 1 mole of oxygen gives 2 moles of water, then 3.125 moles of oxygen will give:

$\frac{2}{1}\times 3.125 mol=6.25 mol$

Mass of 6.25 moles of water :

6.25 mol × 18 g/mol = 112.5 g

Theoretical yield of water = 112.5 g

Experimental yield of water = 75 g

Percentage yield :

$=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{75 g}{112.5 g}\times 100=66.67\%$

The percentage yield of water is 66.67%.

4 0

2 years ago

Solution that is used as the titrant (put in buret) in this experiment. Substance that gives the light pink end point in this ex

Alja [10]

E is the correct answer

6 0

2 years ago

24g of magnesium reacts with 38 g of fluoride o produce __ g magnesium fluoride

Oksana_A [137]

Answer: 62g

Explanation: You add 24g and 38 grams because when a substance reacts with another substance there is the same amount of grams in the product, so there would be 62 grams in the product because that is the amount of grams there is in the reactants.

7 0

2 years ago

Calculate the number of hydrogen atoms in a 110.0 sample of tetraborane(B4H10) . Be sure your answer has a unit symbol if necess

Contact [7]

Answer:

$1.242 \times 10^{25}\text{ atoms H}$

Explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

$\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}$

2. Molecules of B₄H₁₀

$\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}$

3. Atoms of H

$\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}$

8 0

3 years ago