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3 years ago

A straight piece of wire is collided to form a spring physical or chemical change

Chemistry

2 answers:

spayn [35]3 years ago

7 0

Physical because it is still a wire and it has only changed form.

a_sh-v [17]3 years ago

4 0

This would be a physical change.

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Scientists want to eventually collect data from the surface of Venus. Which is a technology that could be used to collect sample

ivanzaharov [21]

Answer:

Explanation:

Because im smart

6 0

3 years ago

Help??

kvasek [131]

Answer:

Mass of sodium chloride decomposed = 24.54 g

Explanation:

Given data:

Mass of sodium chloride decomposed = ?

Mass of chlorine gas formed = 15 g

Solution:

Chemical equation:

2NaCl → 2Na + Cl₂

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 15 g/ 71 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of Cl₂ with NaCl from balance chemical equation.

Cl₂ : NaCl

1 : 2

0.21 : 2×0.21 = 0.42 mol

Mass of Sodium chloride decompose:

Mass = number of moles × molar mass

Mass = 0.42 mol × 58.44 g/mol

Mass = 24.54 g

4 0

2 years ago

Determine the grams of Iron(III) chloride produced if 22.5 g Iron reacts with Chlorine gas. First, balance the chemical equation

TEA [102]

22.5 because it still stays the same

4 0

2 years ago

I need help with this anyone?

valkas [14]

Just look it up on goog^le or a chart

5 0

2 years ago

1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$

Best regards.

4 0

3 years ago