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VARVARA [1.3K]
3 years ago
5

Sex-linked disorders occur more frequently in males than females because males have two X chromosomes and females have two Y chr

omosomes. is this true or false?
Chemistry
2 answers:
Vilka [71]3 years ago
5 0
This is false because males have 1 X and 1 Y chromosome. It's females who have 2 X chromosomes. So, it's false.
zepelin [54]3 years ago
5 0
This is false because a male has an X and a Y chromosome while the female have 2 x chromosomes.
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Which quantity represents 0.500 Mole at STP
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Several ways in which water is an unusual<br> List several ways in<br> substance.
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Boiling Point and Freezing Point., Surface Tension, Heat of Vaporization, and Vapor Pressure, Viscosity and Cohesion, solid State, Liquid State, Gas State.

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What is the basic form for the names of ionic compounds containing a metal that forms more than one type of ion?
professor190 [17]

Answer:

Metal (appropriate charge on metal) nonmetal-ide

Explanation:

1) Write the name of transition metal as shown on the Periodic Table.

2) Write the name and charge for the non-metal.

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3) After the name for the metal, write its charge as a Roman Numeral in parentheses. Example: Iron (II) chloride.

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3 years ago
Cerium (IV) ions are strong oxidizing agents in acid ic solution, oxidizing arsenio us acid to arsen ic acid accor ding to the f
kirill [66]

Answer:

0.2042 M is the original concentration of Ce^{4+} (aq) in the titrating solution.

Explanation:

Mass of As_2O_3 = 0.217 g

Moles of As_2O_3=\frac{0.217 g}{198 g/mol}=0.001096 mol

1 mole of As_2O_3 have 2 mole of As  and 1 mole of H_3AsO_3 have 1 mole of As.

So, from 1 mole of As_2O_3 we will have 2 moles of H_3AsO_3

Then from 0.001096 mol of As_2O_3 :

2\times 0.001096 mol=0.002192 mol of H_3AsO_3

2Ce^{4+}(aq)+H_3AsO_3(aq)+3H_2O(l)\rightarrow 2Ce^{3+}(aq)+H_3AsO_4(aq)+2H^+(aq)

According to reaction, 1 mole of H_3AsO_3 reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of

\frac{2}{1}\times 0.002192 mol=0.004384 mol of cerium (IV) ions.

Volume of the  acidic cerium{IV) sulfate = 21.47 ml =0.02147 L

1 mL = 0.001 L

concentration = \frac{Moles}{Volume(L)}

[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M

0.2042 M is the original concentration of Ce^{4+} (aq) in the titrating solution.

8 0
3 years ago
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