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3 years ago

Explore how the identity of reactants and products can be used to classify types of reactions

1 answer:

8 0

Reactants are substances that start a chemical reaction, and products are substances that are produced in the reaction hope this helps

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Natural remedies are usually more effective and safer because they have thoroughly tested .

mihalych1998 [28]

The answer for this question would be B) False or the second option.

Answer is FALSE: ✅

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3 years ago

Define the term inertia

n200080 [17]

Answer:

Explanation:

Enertia is an integral part of Newton's first law of motion.

It is the tendency of an object to stay at rest or to continue moving until and unless any external unbalanced force, (like, applied force or force of tension or frictional force ) is applied to either move it from rest or change its speed(in other words, accelerate it!!).

Example below, is of ball at rest (fig1) and if this ball is moving straight on a frictionless surface(like ice) it will keep moving!! until, we push it or pull it.

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3 years ago

How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt

Citrus2011 [14]

There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

$(NH_{4})2CO_{3}$

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

$N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g$

We have that the molecular mass of the substance will be:

$MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}$

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

$mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}$

$mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol$

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

$M=\frac{moles_{solute}}{volume_{solution}}$

$M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}$

Now, substituting and calculating, we have:

$volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL$

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

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3 years ago

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Sergio039 [100]

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2 years ago

Assuming that the density of vinegar is 1.005 g/mL, calculate the molatrity of acetic acid in vinegar from an average of 4.6% fo

posledela

Acetic acid is 60.05 grams/mole. In 1 liter of vinegar or 1000 ml there would be 0.046% of acetic acid = 46 ml x 1.005g/ml = 46.23 grams/60.05 grams= 0.77 moles per litre of vinegar.This then would be the concentration of acetic acid in for example 1 liter of vinegar.

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3 years ago