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Nostrana [21]
3 years ago
5

Which of these is a chemical property of a substance

Chemistry
2 answers:
Lady_Fox [76]3 years ago
7 0

Answer:

C. Reactivity

Explanation:

Density, color, melting point, rigidity, fluidity, shape, size etc are all physical properties of any substance.

Hope this helped!

LenaWriter [7]3 years ago
7 0
The answer is D because knowing the melting point is knowing a property
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The arrows below represent the velocities of two vehicles. Which claim explains why the velocity of each vehicle is a vector qua
ollegr [7]

Answer:

It has direction and size.

Explanation:

All physical quantities are generally grouped into two;

i) Vector quantities

ii) Scalar quantities

Vector quantities are those quantities that has both magnitude (size) and direction.

Scalar quantities are those quantities that has magnitude (size) and no direction.

Velocity is a vector quantity because it has direction and size.

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2 years ago
Whats the molar mass of CO
g100num [7]
28.01 g/mol

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7 0
3 years ago
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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
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How can a scientist prevent bias in a scientific investigation
BaLLatris [955]
They must make sure that all things pointed out are fact, not opinion.
5 0
3 years ago
Write the oxidation and reduction half reactions;
luda_lava [24]

Answer:

a)

Fe^{2+}⇒Fe^{3+}+e^-

Br_2+2e^-⇒2Br^-

b)

Mg⇒Mg^{2+}+2e^-

Cr^{3+}+e^-⇒Cr^{3+}

Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

Fe^{2+}⇒Fe^{3+}+e^-

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

Br_2+2e^-⇒2Br^-

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

Mg⇒Mg^{2+}+2e^-

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

Cr^{3+}+e^-⇒Cr^{3+}

3 0
4 years ago
Read 2 more answers
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