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larisa86 [58]
2 years ago
13

Choose the ingredients needed for nuclear fusion. Check all that apply. energy helium gas high temperatures hydrogen gas low pre

ssure
Chemistry
1 answer:
Igoryamba2 years ago
7 0

answer:

high temperatures

hydrogen gas

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A 6.50-g sample of copper metal at 25.0 °c is heated by the addition of 84.0 j of energy. the final temperature of the copper i
anzhelika [568]
The final temperature of the copper is 59.0. The specific heat capacity of copper is 0.38 j/g -k
3 0
2 years ago
How is the separation of dna like a zipper
Sedbober [7]

The two strands must be separated like the two sides of a zipper, by breaking the weak hydrogen bonds that link the paired bases.

<u>Explanation:</u>

  • A double helix structure formed by two polypeptide chains is separated like the two sides of a zipper. A zipper is formed by breaking the weak hydrogen bonds that link the paired bases. During replication, an enzyme "Helicase" travels down the DNA and splits the chain and it forms 2 separate strands.
  • The two DNA strand which has the same sequence must be separated like the two sides of a zipper by breaking weak hydrogen bases. During base pair-rule, the strand are unzipped and each strands is copied.

     

3 0
2 years ago
Which will not appear in the equilibrium constant expression for the reaction below?
n200080 [17]

Answer:

[C] carbon solid

Explanation:

Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.

5 0
2 years ago
James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o
gtnhenbr [62]
This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = <em>57.14%</em>

5 0
2 years ago
Read 2 more answers
What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr
SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, CH_3CH_2CH_3 the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= 1s^{2}2s^{2}2p^{2}

In excited state: 1s^{2}2s^{1}2p^{3}

Hydrogen (atomic number=1): 1s^{1}

All the bonds in the compound is single bond(\sigma-bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in 1s^{1} of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is s-hybridization and the hybridization of Carbon is sp^{3}-hybridization.

The hybridization of each atom is shown in the image.

3 0
2 years ago
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