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nignag [31]
3 years ago
15

Convert 1.75 g/mL to the unit g/m3. Answer Choices: (1.75 x 106 g/m3), (1.75 x 102 g/m3), (1.75 x 10-2 g/m3), (1.75 x 10-6 g/m3)

Chemistry
1 answer:
Marta_Voda [28]3 years ago
4 0
 Answer : 1750000g/cm3
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What is meant by 3p^3
labwork [276]
It means 3p (p being your variable) to the third power, or cubed
7 0
3 years ago
The atomic symbol Superscript 206 subscript 82 upper P b. represents lead-206 (Pb-206), an isotope that has 82 protons and 124 n
Serggg [28]

Answer:

\left \{ {{y=206} \atop {x=82}}Pb \right.

Explanation:

isotopes are various forms of same elements with different atomic number but different mass number.

Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are

  • Alpha particle emission  \left \{ {{y=4} \atop {x=2}}He \right.
  • Beta particle emission    \left \{ {{y=0} \atop {x=-1}}e \right.
  • gamma radiation             \left \{ {{y=0} \atop {x=0}}γ \right.

in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.

Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below

\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right +  \left \{ {{y=0} \atop {x=0}}γ\right.

Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right

8 0
3 years ago
Read 2 more answers
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
PLEASE HURRY I'M TIMED
cricket20 [7]
You should give a picture of the fable!
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