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77julia77 [94]
3 years ago
12

Please help me with this science

Chemistry
1 answer:
MAVERICK [17]3 years ago
6 0

The answer is no. <u>5 grams are equal to 5 milligrams</u>.

"Milli" means million so a <u>milli</u>gram is one-<u>milli</u>onth of a gram.

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The expected first intermediate formed during a halohydrin reaction is:
enot [183]

Answer:

B

Explanation:

The most stable carbonation with OH on the adjacent carbon

5 0
3 years ago
Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.
zaharov [31]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given :

  • V = 2.4 L = 0.0024
  • P = 86126.25 Pa
  • T =  287 K
  • m = 0.622
  • R = 8.314

The ideal gas equation is given below.

n = PV/RT

n = 86126.25 x 0.0024 / 8.314 x 287

n = 0.622 / molar mass (n = Avogardos number)

Molar mass =  7.18 g

Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g

More about the ideal gas equation link is given below.

brainly.com/question/4147359

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4 0
1 year ago
In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white soli
Digiron [165]

Explanation:

When magnesium metal burns is heated i the air it forms magnesium oxide.The balanced chemical reaction is given as:

2Mg+O_2\rightarrow 2MgO

2 moles of magnesium metal when reacts with 1 moles of oxygen it gives 2 moles of magnesium oxide which is white in color.

Some times along with formation of magnesium oxide small amount of magnesium nitride also produced due to which magnesium oxide appears grey in color .The balanced chemical reaction is given as:

3Mg+N_2\rightarrow Mg_3N_2

3 moles of magnesium combines with 1 mol of nitrogen gas to to give 1 mol of magnesium nitride.

6 0
3 years ago
What is the average atomic mass!?
Agata [3.3K]

Answer:

he average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance

6 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
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