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Aleks [24]
2 years ago
11

A student conducts an experiment using a common but potentially harmful bacteria culture. After working with the bacteria, which

is the MOST important safety
practice to follow?
O A. return materials to their proper place
B. wash hands ay tabletops thoroughly
C. tum off any electrical equipment
D. dispose of waste materials
Chemistry
1 answer:
Nadya [2.5K]2 years ago
3 0
I believe the answer is B.
Hope this helps! If not d
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Use the drop-down menus to select the correct name for each of the organic compounds.
Molodets [167]

Answer:

Here's what I get  

Explanation:

CH₃CH₂CH₂CH₂CH₂CH₃ —  hexane

CH₂=CHCH₂CH₂CH₂CH₃ —  hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted

CH₃C≡CCH₃ —  but-2-yne (PIN); 2-butyne is accepted

CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane

CH₃CH₂CHCICH₂CH₃ — 3-chloropentane

5 0
3 years ago
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kap26 [50]
Molar mass of FE2O3=2(55.85)+3(16)=159.7

2.56g*1mol/159.7*2mol/1mol*55.85g/1mol=1.79g
6 0
3 years ago
Balance the equation NH4OH + HCl +H2O= NH3 + Cl2 + H2 + H2O
kompoz [17]
H20 is a simple form of nice
5 0
3 years ago
A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would
Olenka [21]

Answer:

C. Ca_3(PO_4)_2  will precipitate out first

the percentage of Ca^{2+}remaining =  12.86%

Explanation:

Given that:

A solution contains:

[Ca^{2+}] = 0.0440 \ M

[Ag^+] = 0.0940 \ M

From the list of options , Let find the dissociation of Ag_3PO_4

Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}

where;

Solubility product constant Ksp of Ag_3PO_4 is 8.89 \times 10^{-17}

Thus;

Ksp = [Ag^+]^3[PO_4^{3-}]

replacing the known values in order to determine the unknown ; we have :

8.89 \times 10 ^{-17}  = (0.0940)^3[PO_4^{3-}]

\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}  = [PO_4^{3-}]

[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}

[PO_4^{3-}] =1.07 \times 10^{-13}

The dissociation  of Ca_3(PO_4)_2

The solubility product constant of Ca_3(PO_4)_2  is 2.07 \times 10^{-32}

The dissociation of Ca_3(PO_4)_2   is :

Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}

Thus;

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33} = (0.0440)^3  [PO_4^{3-}]^2

\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}=   [PO_4^{3-}]^2

[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}

[PO_4^{3-}]^2 = 2.43 \times 10^{-29}

[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}

[PO_4^{3-}] =4.93 \times 10^{-15}

Thus; the phosphate anion needed for precipitation is smaller i.e 4.93 \times 10^{-15} in Ca_3(PO_4)_2 than  in  Ag_3PO_4  1.07 \times 10^{-13}

Therefore:

Ca_3(PO_4)_2  will precipitate out first

To determine the concentration of [Ca^+] when  the second cation starts to precipitate ; we have :

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33}  = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2

[Ca^{2+}]^3 =  \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}

[Ca^{2+}]^3 =1.808 \times 10^{-7}

[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}

[Ca^{2+}] =0.00566

This implies that when the second  cation starts to precipitate ; the  concentration of [Ca^{2+}] in the solution is  0.00566

Therefore;

the percentage of Ca^{2+}  remaining = concentration remaining/initial concentration × 100%

the percentage of Ca^{2+} remaining = 0.00566/0.0440  × 100%

the percentage of Ca^{2+} remaining = 0.1286 × 100%

the percentage of Ca^{2+}remaining =  12.86%

5 0
3 years ago
What happens when the concentration of water inside a cell is lower than the concentration of water outside the cell?
adell [148]

Answer:

Water outside the cell will flow inwards by osmosis to attain equilibrium

Explanation:

In the hypotonic environment, the concentration of water is greater outside the cell and the concentration of solute is higher inside. A solution outside of a cell has a lower concentration of solutes relative to the cytosol.

If concentrations of dissolved solutes are greater inside the cell, the concentration of water inside the cell is correspondingly lower. As a result, water outside the cell will flow inwards by osmosis to attain equilibrium.

Osmosis is a process by which molecules of a solvent tend to pass from a less concentrated solution into a more concentrated one through a semipermeable membrane.

4 0
2 years ago
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