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3 years ago

When you strike a match you produce frictional heat energy what does this input of energy present in a chemical reaction

Chemistry

2 answers:

Talja [164]3 years ago

4 0

Answer:

If this is that MC question then the answer is D.

Explanation:

The correct answer is D. The energy needed for a chemical reaction is called the activation energy. This energy is to be present in order for the reaction to happen. This can be achieved by supplying or releasing heat from a system of reactants. This can achieved faster by using a catalyst.

evablogger [386]3 years ago

3 0

Answer:

Activation energy

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Given the following chemical reaction:

True [87]

Answer:

Explanation:

one volume of nitrogen to react

5 0

2 years ago

You used a sample of 20 particles to determine the average mass. Do you think the average mass you calculated depends on the num

Vedmedyk [2.9K]

No because its not the proper mass.

4 0

3 years ago

Calculate the moles and grams of solute in 2.0 L of 0.30M Na SO..

Ostrovityanka [42]

Answer:

Number of moles of solute = 0.6 mole

Mass =13.8 g

Explanation:

Given data:

Number of moles of sodium = ?

Volume = 2.0 L

Molarity = 0.30 M

Mass in gram of sodium= ?

Solution:

Number of moles:

Molarity = number of moles of solute / volume in litter

Number of moles of solute = Molarity × volume in litter

Number of moles of solute = 0.30 M × 2.0 L

Number of moles of solute = 0.6 mole

Mass in gram:

Mass = Number of moles × molar mass

Mass = 0.6 mole× 23 g/mol

Mass =13.8 g

3 0

3 years ago

What is the relationship between a compound and the elements it is made from?

dalvyx [7]

The compound may have properties that are very different from those of the elements it is composed from.

6 0

2 years ago

If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L

Tom [10]

Answer:

Approximately $2.53\; \rm L$ (rounded to three significant figures) assuming that ${\rm HCl}\, (aq)$ is in excess.

Explanation:

When ${\rm HCl} \, (aq)$ and ${\rm AgNO_3}\, (aq)$ precipitate, ${\rm AgCl} \, (s)$ (the said precipitate) and $\rm HNO_3\, (aq)$ are produced:

${\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)$ (verify that this equation is indeed balanced.)

Look up the relative atomic mass of $\rm Ag$ and $\rm Cl$ on a modern periodic table:

$\rm Ag$ : $107.868$ .
$\rm Cl$ : $35.45$ .

Calculate the formula mass of the precipitate, $\rm AgCl$ :

$\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of $\rm AgCl$ formula units in $45.00\; \rm g$ of this compound:

$\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}$ .

Notice that in the balanced equation for this reaction, the coefficients of ${\rm AgNO_3} \, (aq)$ and ${\rm AgCl}\, (s)$ are both one.

In other words, if ${\rm HCl}\, (aq)$ (the other reactant) is in excess, it would take exactly $1\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $1\; \rm mol \!$ of ${\rm AgCl}\, (s)\!$ formula units.

Hence, it would take $0.313987\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $0.313987\; \rm mol\!$ of ${\rm AgCl}\, (s)\!$ formula units.

Calculate the volume of the ${\rm AgNO_3} \, (aq)\!$ solution given that the concentration of the solution is $0.124\; \rm mol \cdot L^{-1}$ :

$\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}$ .

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of ${\rm AgNO_3} \, (aq)\!$ .)

In other words, approximately $2.53\; \rm L$ of that ${\rm AgNO_3} \, (aq)\!$ solution would be required.

4 0

3 years ago