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Anettt [7]
2 years ago
6

I need help with a DNA research paper

Chemistry
2 answers:
quester [9]2 years ago
5 0

Answer:

XD ok wheres the paper

Explanation:

JulsSmile [24]2 years ago
3 0
Let me see the paper
I’m good with DNA hehehehe
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When an electron loses electrons it becomes more 1 negative. 2 none of the above. 3 neutral 4 positive.
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The answer would be more positive
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4 years ago
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Place elements in each in set of decreasing atomic radius: O,S,Si
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Si has the largest, then S, then O
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CCC Structure and Function Why is the arrangement of the constituent atoms or
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Since the atoms or molecules in a solid do not translate hence, the structures of solids usually described in terms of the positions  of the constituent atoms rather than their motion.

<h3>States of matter</h3>

There are three state of matter which are;

  • Solid
  • Liquid
  • Gas

In a solid, atoms or molecules retain their positions and do not translate, hence, the structure of atoms or molecules are most important in determining the structure of a solid.

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Learn more about the states of matter: brainly.com/question/9402776

8 0
3 years ago
Balance the following equation and find molar ratios from the equation,
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To balance it, it would be N2 + 3H2 ------> 2NH3. 

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7 0
3 years ago
You have 15 grams of ice at 0°C. How much energy is needed to melt the ice and warm it to 25°C. Ice melts at 0°C. The specific h
azamat

Answer:

d. 1600 calories

Explanation:

The heat of fusion of water, L, is the amount of heat per gram required to melt the ice to water, a process which takes place at a constant temperature of 0 °C. The specific heat of water, c, is the amount of heat required to change the temperature of 1 gram of water by 1 degree Celsius.

We will convert the units of c from Jg⁻¹°C⁻¹ to cal·g⁻¹°C⁻¹ since the answers are provided in calories. The conversion factor is 4.18 J/cal.

(4.18 Jg⁻¹°C⁻¹)(cal/4.18J) = 1 cal·g⁻¹°C⁻¹

First we calculate the heat required to melt the ice, where M is the mass:

Q = ML = (15 g)(80 cal/g) = 1200 cal

Then, we calculate the heat required to raise the temperature of water from 0 °C to 25 °C.

Q = mcΔt = (15 g)(1 cal·g⁻¹°C⁻¹)(25 °C - 0 °C) = 380 cal

The answer is rounded so that there are two significant figures

The total heat required for this process is (1200 cal + 380 cal) = 1580 cal

The rounded answer is 1600 calories.

3 0
3 years ago
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