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Bad White [126]
3 years ago
10

What is the full meaning of DR.HERC.​

Chemistry
1 answer:
n200080 [17]3 years ago
5 0

Answer:

It is

D- <u><em>DEFINE</em></u> the problem

R-<em><u>RESEARCH</u></em> on the problem

H- Carry out a <u><em>HYPOTHESIS</em></u>

E- carry out an <em><u>EXPERIMENT</u></em>.

R-Analyse the <u><em>RESULT</em></u>

C-summarise the <u><em>CONCLUSION</em></u>.

Explanation:

Hope it helps.

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What is the part of the earth that includes the solid crust and the rigid uppermost part of the martle
faltersainse [42]

Answer:

I can't understand

Explanation:

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6 0
3 years ago
A 1.2 kg block of iron at 32 ∘C is rapidly heated by a torch such that 12 kJ is transferred to it. What temperature would the bl
lbvjy [14]

Answer:

For iron

Final temperature = 54,22°C

For copper

Final Temperature = 63.67 °C

Explanation

Hello,

You are using a torch to warm up a block of iron that has an initial temperature of 32°C.

The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.

So you need to use the heat capacity equation as follow in the paper.

The equation has to have all terms in the same units, so:

q = 12000 J

s = 0.450 J / g °C

m = 1200 g

Ti = 32 °C

Download odt
3 0
3 years ago
One phosphorus, three chlorine, one oxygen
4vir4ik [10]
Phosphoryl chloride :

POCl3
<span>
hope this helps!.

</span>
5 0
3 years ago
What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
7 0
3 years ago
How many significant figures does .150, 20.20, and 25.00 have?
viva [34]
.150 = 3
20.20 = 4
25.00 = 4
If a 0 does not have a number or a period after it, it is not significant.
If the 0 is behind the decimal point, it is always significant.
6 0
3 years ago
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