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3 years ago

Explain why your hair might stand up when going down a slide.

Chemistry

2 answers:

elena55 [62]3 years ago

8 0

Answer:

it is like when rubbing a balloon on your head the energy from your body and from you moving the balloon fast makes energy to make your hair stand up.

Explanation:

Hope this helps :)

nikitadnepr [17]3 years ago

6 0

Answer:

the electricity

Explanation:

objects with the same charge repel each other. Because they have the same charge, your hair will stand on end. Your hairs are simply trying to get as far away from each other as possible

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___ are in group 18

Irina-Kira [14]

Answer:

<h3>a. noble gases</h3>

Explanation:

Groups are numbered 1–18 from left to right. The elements in group 1 are known as the alkali metals; those in group 2 are the alkaline earth metals; those in 15 are the pnictogens; those in 16 are the chalcogens; those in 17 are the halogens; and those in 18 are the noble gases.

Hope this helps

8 0

3 years ago

In the reaction, Zn(s) + 2 HCl(aq) --> ZnCl2 (aq)+ H2(g), 25 grams of Zn are reacted with 17.5 g of HCl. How many grams of H2

Gennadij [26K]

Answer:

0.480 g of H₂ are produced, in the reaction.

Explanation:

This is the reaction:

Zn(s) + 2HCl → ZnCl₂ (aq) + H₂ (g)

We havethe mass of both reactants, so we must work with them to find out the limiting reactant and then, determine the amount of H₂ produced.

Let's convert the mass to moles ( mass / molar mass)

25 g / 65.41 g/mol = 0.382 moles Zn

17.5 g / 36.45 g/mol = 0.480 moles HCl

Ratio is 1:2, so 1 mol of Zn react with the double of moles of HCl.

0.382 moles of Zn would need the double of moles to react, so (0.382 .2) = 0.764 moles of HCl. → We only have 0.480 moles, so the acid is the limiting.

Now let's determine the moles of H₂ formed.

Ratio is 2:1, so If i take account the moles I have, I will produce the half of moles of my product.

0.480 moles / 2 = 0.240 moles of H₂ are produced.

To find out the mass, we must multiply mol . molar mass

0.240 mol . 2g/mol = 0.480 g

3 0

3 years ago

Use the information from the diagram to calculate the enthalpy of combustion for methane. (HELP ASAPPP!!)

Blizzard [7]

Answer:

where is the diagram sisters

6 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$