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3 years ago

What is one practical use of the paper battery

Chemistry

1 answer:

love history [14]3 years ago

5 0

Answer:

A paper battery can work even if it is folded, cut or rolled up. A Paper battery consists mainly of carbon and paper; it can be used to power pacemakers within the body. A paper battery can be used both as a capacitor and battery. It is an ultra-thin storage device.Oct 10, 2014

Explanation:

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______ is the use of enzymes is necessary to increase the activation energy requirements of a chemical reaction.

Masja [62]

Answer:

Catalysts

Explanation:

Catalysts lower the aviation energy.

6 0

3 years ago

What is the concentration of h+ ions in a 2.20 m solution of hno3?

kvv77 [185]

The question is improperly formatted.

What is the concentration of H+ ions in a 2.2 M solution of HNO3.

Answer:-

2.2 moles of H+ per litre

Explanation:-

M stands for molarity. 2.2 M means 2.2 moles of HNO3 is present per litre of the solution.

Now HNO3 has just 1 H in it's formula. HNO3 would give H+. So 2.2 moles of HNO3 would mean 2.2 moles of H+ per litre.

7 0

4 years ago

_______ is the application of scientific knowledge to solve problems or create new products.

astraxan [27]

Having resources? i guess...
maybe it is critical thinking?

3 0

3 years ago

Read 2 more answers

what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c

slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

$Q = m \times C \times \bigtriangleup T$

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

$C = \frac{Q}{m \times \bigtriangleup T}$

Let us plug in the values in above equation.

$C = \frac{300J}{267g \times 12 C}$

$C = \frac{300J}{3204 g C}$

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0

3 years ago

A 51.24-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be d

g100num [7]

Answer:

0.40 L

Explanation:

Calculation of the moles of $Ba(OH)_2$ as:-

Mass = 51.24 g

Molar mass of $Ba(OH)_2$ = 171.34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{51.24\ g}{171.34\ g/mol}$

$Moles= 0.2991\ mol$

Volume = 1.20 L

The expression for the molarity is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M$

Thus,

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$Molarity_{working\ solution}=0.100\ M$

$Volume_{working\ solution}=1\ L$

$Volume_{stock\ solution}=?$

$Molarity_{stock\ solution}=0.24925\ M$

So,

$0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}$

$Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L$

<u>The volume of 0.24925M stock solution added = 0.40 L </u>

6 0

3 years ago