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3 years ago

how many moles of solute are contained in 500.00 milliliter solution that has a concentration of 0.35M

Chemistry

1 answer:

igomit [66]3 years ago

6 0

1.75 moles of Solute or 1.75n.

To find this, you'll need to use the molarity equation which is M = $\frac{n}{v}$

List the given items to organize better:

M (Molarity) = .35

V (Volume) = 5L [must always be in Liters]

n (Moles) = ?

Now, for the equation, substitute the variables with the given numbers:

.35 = $\frac{n}{5}$

Isolate the variable by multiplying the volume, 5, onto both sides. This will give you the answer of:

1.75 = n

Hope this helps!

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The pH of a vinegar solution is 4.15. What is the H+ concentration of the solution

marishachu [46]

The pH of a vinegar solution is 4.15. To find the H+ concentration of the solution use the following equation -log(H+)=pH. Insert the pH into the equation to get, -log(H+) = 4.15 Rearrange the equation to get, 10^(-4.15) = H+ Finally, you can solve for H+. The hydrogen ion concentration of the vinegar solution is .0000708 M.

3 0

3 years ago

Read 2 more answers

Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati

worty [1.4K]

Answer:

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4 + Cl-

Firstly, we calculate the number of moles of the ammonia as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

NH4+ H2O ⇄ NH3 H3O+

I 0.242 0 0

C -X +x +X

E 0.242-X X X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b = K w

, where K w- the self-ionization constant of water, equal to

10 ^-14 at room temperature

This means that you have

K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈ 0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0

3 years ago

What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The

Effectus [21]

Answer: The energy released in the given nuclear reaction is 1.3106 MeV.

Explanation:

For the given nuclear reaction:

$_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{40}\textrm{K}$ = 39.963998 u

Mass of $_{20}^{40}\textrm{Ca}$ = 39.962591 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(39.963998-39.962591)=0.001407u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.001407u)\times c^2$

$E=(0.001407u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=1.3106MeV$

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0

3 years ago

If we react 30 grams of ethane (C2H6 ) with 320 grams of oxygen, how many grams of carbon dioxide would we make

V125BC [204]

The mass of carbon dioxide that would be made by reacting 30 grams C2H6 with 320 grams O2 will be 80 grams

From the balanced equation of the reaction:

$2C_2H_6+7O_2---4CO_2+6H_2O$

The mole ratio of C2H6 to O2 is 2:7.

Mole of 30 grams C2H6 = mass/molar mass

= 30/30

= 1 mole

Mole of 320 grams O2 = 320/32

= 10 moles

Thus, C2H6 is the limiting reactant.

Mole ratio of C2H6 to CO2 according to the equation = 1:2

Since the mole of C2H6 is 1, the equivalent mole of CO2 would, therefore, be 2.

Mass of 2 moles CO2 = mole x molar mass

= 2 x 44

= 88 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886?referrer=searchResults

6 0

3 years ago

If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera

Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0

3 years ago