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2 years ago

11

Which chemical equation is unbalanced? c o2 right arrow. co2 sr o2 right arrow. 2sro 6h2 3o2 right arrow. 6h2o h2 h2 o2 right ar

row. h2o h2o

1 answer:

netineya [11]2 years ago

6 0

The unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical equation is one in wgich the moles of atoms in the reactants side is equal to the moles of atoms on the product side.

The given equations of reaction is not clearly stated.

Therefore, the unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

Learn more about balanced equations at: brainly.com/question/11904811

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Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

GarryVolchara [31]

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

6 0

3 years ago

What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?

insens350 [35]

First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm

6 0

3 years ago

How can scientists use stoichiometry to ensure that commercial products contain the components that manufacturers claim they do?

Rina8888 [55]

Answer:

I know you have been waiting awhile for this question to be answered :)

Stoichiometry is used in industry quite often to determine the amount of materials required to produce the desired amount of products in a given useful equation. Each one of these products requires stoichiometry. There would be no products from these industries without chemical stoichiometry.

Explanation:

Hopefully this helps :D

Sorry you had to wait so long :(

4 0

3 years ago

Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL

dybincka [34]

Answer:

$M_2=3.34x10^{-3}M$

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

$M_1V_1=M_2V_2$

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

$M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M$

Best regards!

5 0

3 years ago

Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.

SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

HClO ⇄ ClO- + H+

start: 0.05 0 0

change -x +x +x

balance 0.05-x x x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

$Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}$

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

$3x10^{-8}=\frac{x^{2} }{0.05}$

clearing the x and calculating its value we have:

$x=3.87x10^{-5}=[H+]=[ClO-]$

the pH can be calculated by:

$pH=-log[H+]=-log[3.87x10^{-5}]=4.41$

7 0

3 years ago