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3 years ago

5

Predict whether or not a precipitate forms upon mixing 175.0 ml of a 0.0055 mkcl solution with 145.0 ml of a 0.0015 m agno3 solu

tion. identify the precipitate, if any.

1 answer:

BabaBlast [244]3 years ago

8 0

There would be no precipitate unless there is an aqueous solution (aq) or water (w) identified in the question... I think.

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Which unit is not a measurement of volume?<br><br> cm<br> L<br> ml<br> Cm^3

Mademuasel [1]

Centimeters are units of length.

8 0

3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0

3 years ago

1.What is another term for the 24 hour clock?

s344n2d4d5 [400]

Answer:

military time

Explanation:

military time

5 0

2 years ago

A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio

kondor19780726 [428]

Answer:

The final volume of the cylinder is 1.67 L

Explanation:

Step 1: Data given

Initial volume = 0.250 L

external pressure = 2.00 atm

Expansion does 288 J of work on the surroundings

Step 2: Definition of reversible work:

Wrev = -P(V2-V1) = -288 J

The gas did work, so V2>V1 (volume expands) and the work has a negative sign.(Wrev<0)

V2 = (-Wrev/P) + V1

⇒ with Wrev = reverse work (in J)

⇒ with P = the external pressure (in atm)

⇒ with V1 = the initial volume

We can see that your pressure is in atm and energy in J

To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :

R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol

V2 =- (-288 J * (0.08206 L*atm/K*mol /8.314 J/mol *K))/2.00 atm + 0.250L

V2 = 1.67 L

The final volume of the cylinder is 1.67 L

8 0

2 years ago

A piece of metal that has a density of 5.2 g/cm3 and a mass of 100 g was placed in a full jar of water. How many mL of water was

Klio2033 [76]

Answer:

100g / (5.2g/cm3)

= 100g / (5.2g / 1cm3)

= 100g x 1cm3 / 5.2 g

= 19.2 cm3

Since 1 cm3 = 1 ml, your answer is 19.2 ml.

5 0

3 years ago