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insens350 [35]
3 years ago
13

Cavendish carried out a second experiment to test the properties of the hydrogen gas produced in the experiment. Was this experi

ment also a chemical reaction? Explain.
Chemistry
2 answers:
Arada [10]3 years ago
7 0

Answer:

Yes, this experiment is also an example of a chemical reaction. The original substance, hydrogen, disappeared, and a new substance, water, was formed.

Explanation:

love history [14]3 years ago
6 0

Yes, this experiment is also an example of a chemical reaction. The original substance, hydrogen, disappeared, and a new substance, water, was formed.

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24. The air in a small 250 cm birthday balloon is at a pressure of 760 torr. A boy sits on it at
kvv77 [185]

Answer:

1.14atm

Explanation:

Given parameters:

V1 = 250cm³ ;

              1000cm³ = 1dm³;  so this is 0.25dm³

P1  = 760torr

            760torr  = 1atm

       

V2  = 220cm³ ; 0.22dm³

Unknown:

New pressure = ?

Solution:

To solve this problem, we apply Boyle's law and we use the expression below:

       P1 V1 = P2V2

The unknown is P2;

            1 x 0.25  = P2 x 0.22

                 P2  = 1.14atm

7 0
3 years ago
Which of the following best represents potential energy being converted to kinetic energy?
Rufina [12.5K]
I believe the answer is A
5 0
3 years ago
Read 2 more answers
Firewood burns and ashes remain<br><br> is it a chemical or physical change?
erastova [34]
It is a physical change because you can not put it back like it was
4 0
3 years ago
Read 2 more answers
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
3 years ago
(06.04 HC)
Ksju [112]

The mass of sodium chloride at the two parts are mathematically given as

  • m=10,688.18g
  • mass of Nacl(m)=39.15g

<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>

Generally, the equation for ideal gas is mathematically given as

PV=nRT

Where the chemical equation is

F2 + 2NaCl → Cl2 + 2NaF

Therefore

1.50x15=m/M *(1.50*0.0821)

1-50 x 15=m/58.5 *(1.50*0.0821)

m=10,688.18g

Part 2

PV=m'/MRT

1*15=m'/58.5*0.0821*273

m'=39.15g

mass of Nacl(m)=m'=39.15g

Read more about Chemical Reaction

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7 0
2 years ago
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