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3 years ago

Magbigay ng 3 halimbawa ng lipunang sibil. Paki sagot ng wasto plies >_<

Chemistry

1 answer:

Sliva [168]3 years ago

6 0

Answer:

i not anther stand your bla bla bla

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Calculate the hydronium ion and hydroxide ion concentrations in human blood at 25oc

sukhopar [10]

Using the equation, pH = − log [H+] , we can solve for [H+] as,

− pH = log [H+] ,

[H+] = 10−pH.

Exponentiate both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,

[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.

5 0

3 years ago

Some chemical factories make hydrogen gas (H2) using a process called steam methane reforming. In this process, methane (CH4) co

erica [24]

Answer:

steam and methane

Explanation:

5 0

3 years ago

An electron that is in the highest energy level of an atom and determines the atom's chemical properties is called a(n)

ss7ja [257]

The answer is C: Valence electron

3 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago

If element x has 72 protons, how many electrons does it have? electrons

saw5 [17]

Elements have neutral charges such as Na so if X has 72 protons + charges then it must also have 72 electrons with - charges

hope that helps

6 0

3 years ago

Magbigay ng 3 halimbawa ng lipunang sibil. Paki sagot ng wasto plies >_<​

Magbigay ng 3 halimbawa ng lipunang sibil. Paki sagot ng wasto plies >_<