All categories

2 years ago

A solution with a ph of 7 must be _____. an acid a base neutral water

Chemistry

2 answers:

steposvetlana [31]2 years ago

6 0

Answer: A solution with a pH of 7 must be neutral.

Explanation:

If a solution has pH of 7 then the concentration of $H^+$ ions and $OH^{-}$ ions will be equal.

Thus, the concentration for both $H^+$ ions and $OH^{-}$ ions will be $1 \times 10^{-7}$ . Generally, pH at 7 is considered as neutral.

Thus, it is concluded that a solution with pH of 7 must be neutral.

Setler [38]2 years ago

3 0

Neutral because an acid is between 1 and 6 and a base is between 8 and 14

You might be interested in

A chemist sets up a chemical reaction but finds that none of the reactant molecules have a required activation energy. What is t

Anvisha [2.4K]

Activation energy is the minimum amount of energy that the colliding reactant molecules must possess for the formation of products. Lower the activation energy, higher will be chance of formation of products. So activation energy is the minimum energy requirement that has to overcome for the reaction to be completed. Therefore, when in a chemical reaction the reactant molecules do not collide with required activation energy, the collisions will not be fruitful even if they are properly oriented which means that the products will not form.

Hence the correct answer will be B.) no products will be formed

5 0

3 years ago

Read 2 more answers

Determine the mass fraction of iron in its compounds:

Shkiper50 [21]

Explanation:

We have a molecule composed of 3" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; one; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">33 iron atoms, and 4" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: p; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">44 atoms of another element. We are given the following information: it has 2.36 g" role="presentation" style="box-sizing: inherit; margin: 0px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">2.36 g2.36 g of iron for 3.26 g" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: -wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">3.26 g3.26 g of molecule.

I want to find the molar mass of the compound, I have tried so far:

m=3.26 g=0.00326 kg" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; font-family: inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">m=3.26 g=0.00326 kgm=3.26 g=0.00326 kg

Since it has 3" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; nt-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; font-family: inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">FeFe and 4" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; ; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">44 atoms of an unknown substance, therefore:

3+4=7 atoms,1 mol=6.022⋅1023 atoms76.022⋅1023=1.16⋅10−23" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: : ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; width: 10000em !important; ; font-family: inherit; eight: none; min-width: 0px; min-height: 0px; position: relative;">0.003260.00326 by 1.16⋅10−23" role="presentation"; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">1.16⋅10−231.16⋅10−23 and I obtained 2.79429⋅1019" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; f inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">2.79429⋅10

IT'S TOTAL ANSWER OF ITS AND THIS QUESTION IS IN MATHEMATION FINAL EXAM. PLEASE GIVE❤ AND MARK ME A BRAINLIST

7 0

2 years ago

How many moles are there in 3.45 moles of Carbon

VladimirAG [237]

Answer:

It would be 151.832775 because one mole is 44.0095*3.45 i hope this helps!

Explanation:

7 0

3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

3 years ago

/msk-yoxd-wmh<br> cya there

oksano4ka [1.4K]

Answer:

aaaaahhhhhhhhhhhhhhhhhhhhhhhhhh

Explanation:

6 0

2 years ago