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ankoles [38]
3 years ago
10

What part of dalton’s atomic theory is disproved by the existence of isotopes?

Chemistry
1 answer:
shepuryov [24]3 years ago
3 0
The part that says all atoms of a specific element always have the same mass. Isotopes of an atom have varying masses compared to the standard version of the element.
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What kingdom do fungi belong
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8 0
3 years ago
In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu
Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

7 0
2 years ago
"A crosslinked copolymer consists of 57 wt% ethylene (C2H4) repeat units and 43 wt% propylene (C3H6) repeat units. Determine the
ch4aika [34]

Explanation:

Percentage ethylene by weight = 57%

Percentage propylene by weight = 43%

Suppose in 100 grams of polymer:

Mass of ethylene = 57 g

Mass of propylene = 43 g

Moles of ethylene = \frac{57 g}{28 g/mol}=2.036 mol

Moles of propylene = \frac{43g}{42g/mol}=1.024 mol

1 mole = N_A =6.022\times 10^{23} molecules/ atoms

Units of ethylene = 2.036 mol\times N_A

Units of propylene = 1.024 mol\times N_A

a) Fraction of ethylene units:

=\frac{2.036 mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{509}{765}

b ) Fraction of propylene units:

=\frac{1.024mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{256}{765}

3 0
3 years ago
A chemist encounters an unknown metal. They drop the metal into a graduated cylinder containing water, and find the volume chang
Mashutka [201]

Answer:

The density of the metal is 0.561 g/mL

Explanation:

The computation of the density of the metal is shown below;

As we know that

The Density of the metal is

= \frac{mass}{volume}

where,

Mass = 4.9g

Change in volume = 6.9 mL

Now place these values to the above formula

So, the density of the metal is

= \frac{4.9g}{6.9mL}

= 0.561 g/mL

Hence, the density of the metal is 0.561 g/mL

We simply applied the above formula so that the correct density could arrive

5 0
2 years ago
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