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3 years ago

When a chemical reaction occurs with only one substance, what happens to the atoms of that substance?

2 answers:

7 0

Answer:when a chemical reaction occurs with only one substance, the atoms of that substance rearranges to form new bonds to make the products.

Explanation: During chemical reaction, the bonds between the reactants are broken and the atom bonds in new ways to form products that are chemically and physically different from the reactants.

Example is the breakdown reaction of hydrogen peroxide(the reactant) into water and oxygen( products). Here, 2 molecules of hydrogen peroxide is broken down to form 2 molecules of water and one molecule of oxygen. The atoms that started out in the hydrogen peroxide molecules are rearranged to form water molecules and oxygen molecules.

miskamm [114]3 years ago

6 0

Answer:

The atoms of the substance are rearranged

Explanation:

When a chemical reaction involves only one substance, the atoms of the reactant substance rearranges to yield the product substance(s).

Chemical bonds in the reactant substance are broken and new bonds are formed as the atoms in the reactant substance forms the product substance (s).

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To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what

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A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin

anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

$\Delta L = L_0 \alpha \Delta T$

Where,

$L_0 =$ Initial Length

$\alpha =$ Thermal expansion coefficient

$\Delta T =$ Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

$0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C$

$103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C$

The coefficient of thermal expansion given is

$\alpha = 1.1*10^{-5}/\°C$

The initial length would be,

$L_0 = 565m$

Replacing we have to,

$\Delta L = L_0 \alpha \Delta T$

$\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))$

$\Delta L = 0.355m$

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3 years ago

The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.

Pepsi [2]

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = $[tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}$ [/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = $\frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}$

E = 1.655 x 10⁷ N/C

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