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3 years ago

Identify the independent /manipulated variable

Chemistry

1 answer:

fiasKO [112]3 years ago

7 0

Answer:

independent- test variable

manipulated - outcome variable; if needed, there could be changes.

Explanation:

knowledge

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SOMEONE PLEASE HELP

ratelena [41]

Ok what do u need help with

4 0

3 years ago

Read 2 more answers

If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o

sweet-ann [11.9K]

Answer:

The molarity of the $AgNO_3$ solution is $4.02 \times 10^4 M$

Explanation:

The Balanced chemical equation is

$1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)$

Mole ratio of $AgNO_3$ : KCl is 1 : 1

So moles $AgNO_3$ = moles KCl

$Moles KCl = \frac {mass}{molarmass}$

$= \frac {0.785 mg}{(39.1+35.5 g per mol)}$

$= \frac {0.000785 g}{74.6 g per mol}$

$= 0. 0000105 mol KCl$

$= 0.0000105 mol AgNO_3$

So Molarity

$= \frac {moles of solute}{(volume of solution in L)}$

$= \frac {0.0000105 mol}{26.2 mL}$

$=\frac {0.0000105 mol}{0.0262 L}$

= 0.000402M or mol/L is the Answer

(Or) $4.02 \times 10^4 M$ is the Answer

6 0

3 years ago

When the following oxidation-reduction occurs, what is the balanced reduction half-reaction after the electrons in both half rea

Lostsunrise [7]

Answer : The balanced reduction half-reaction is:

$3Cu^{2+}+6e^-\rightarrow 3Cu$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The given balanced redox reaction is :

$Al(s)+Cu^{2+}(aq)\rightarrow Al^{3+}(aq)+Cu(s)$

The half oxidation-reduction reactions are:

Oxidation reaction : $Al\rightarrow Al^{3+}+3e^-$

Reduction reaction : $Cu^{2+}+2e^-\rightarrow Cu$

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

Oxidation reaction : $2Al\rightarrow 2Al^{3+}+6e^-$

Reduction reaction : $3Cu^{2+}+6e^-\rightarrow 3Cu$

The balanced redox reaction will be:

$2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)$

Thus, the balanced reduction half-reaction is:

$3Cu^{2+}+6e^-\rightarrow 3Cu$

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3 years ago

Give 3 property's of an apple (density)

erastovalidia [21]

Answer:

It's a fruit. It's red, yellow or green. It's edible.

Explanation:

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3 years ago

What is the sum of all charges in the formula for an ionic compound?

Inessa05 [86]

Answer:

The sum of all the charges should equal the charge on the ion.

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