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crimeas [40]
2 years ago
5

Food is heated up in a microwave. The heat transfer used in a microwave is

Chemistry
2 answers:
Vikentia [17]2 years ago
7 0
D.) radiation because of the heat waves
azamat2 years ago
6 0
Radiation 

hope that helped!
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If the experiment were repeated using a cuvettes with a path length of 2.0 cm, how would that affect the absorbance values for t
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Answer:

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How can we reduce the production of harmful gases that form acid rain
oksano4ka [1.4K]

Answer:

A great way to reduce acid rain is to produce energy without using fossil fuels. Instead, people can use renewable energy sources, such as solar and wind power. Renewable energy sources help reduce acid rain because they produce much less pollution.

7 0
3 years ago
A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 da
Flura [38]
55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
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7 0
3 years ago
Read 2 more answers
Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.
tamaranim1 [39]

Answer:

1500kg/m^3

Explanation:

Formula:

d=m/v

Given:

m=1.5kg

v=1000cm^3

(The side length of a cube is always equal to the others)

Required:

d=?

Solution:

d=m/v

d=1.5kg/1000cm^3

d=1.5kg/0.001m^3

d=1500kg/m^3

Hope this helps ;) ❤❤❤

3 0
3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

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

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
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