Taro is a _______ meaning that it was brought to the hawaiian islands by boat

0.05 grams od cobalt ii oxalate will dissolve in one liter of solution at 25 degree celcius. Calculate the ksp for cobalt ii oxa

Veseljchak [2.6K]

Answer:

The equilibrium expression is:

CoC2O4(s)⇌Co2+(aq)+C2O2−4(aq)

For this reaction:

Ksp = [Co2+][C2O2−4]=1.96×10−8

Explanation:

Batteries will not clot if cobalt ions are removed from its cells. Some blood collection tubes contain salts of the oxalate ion,

C2O2−4

, for this purpose. At sufficiently high concentrations, the calcium

and oxalate ions form solid, CoC2O4·H2O (which also contains water bound in the solid). The concentration of Co2+ in a sample of blood serum is 2.2 × 10–3M. What concentration of

C2O2−4

ion must be established before CoC2O4·H2O begins to precipitate.

CoC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CoC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Co2+], we can solve for the concentration of

C2O2−4

that is necessary to produce the first trace of solid:

5 0

3 years ago

A sample of water with a mass of 27.56g and an unknown temperature loses 2443 Joules. If the final temperature is found to be 62

mixer [17]

Answer:

41.3 °C

Explanation:

From the question given above, the following data were obtained:

Mass (M) of water = 27.56 g

Heat (Q) loss = 2443 J

Final temperature (T2) = 62.5 °C

Initial temperature (T1) =?

NOTE: The specific heat capacity (C) of water is 4.18 J/g°C

Thus, we can obtain the initial temperature of the water by using the following formula:

Q = MC(T2 – T1)

2443 = 27.56 × 4.18 (62.5 – T1)

2443 = 115.2008 (62.5 – T1)

Divide both side by 115.2008

2443 / 115.2008 = (62.5 – T1)

21.20645 = 62.5 – T1

Collect like terms

21.20645 – 62.5 = – T1

– 41.3 = – T1

Divide both side by – 1

– 41.3 /– 1= – T1 / –1

41.3 = T1

T1 = 41.3 °C

Thus, the initial temperature of the water was 41.3 °C

8 0

3 years ago

HELP ASAP!!!!

never [62]

Answer:

what are the answer choices??

7 0

3 years ago

Upon landing, the 90.7 kg carbon fiber brakes of an airliner heat up 312∘C, producing heat. As the brakes start to cool back to

Crank

Given,

Mass of carbon fiber brakes = 90.7 kg = 90.7 x 1000 = 90700 g

Mass of the rubber tires = 123 kg = 123 x 1000 = 123000

Specific heat of carbon fiber = 1.400Jg∘C

As all the heat is transferred from the carbon fiber brakes to the rubber tires, the heat lost by the carbon fiber brakes is equal to the heat gained by the rubber tires.

So the temperature change of carbon fiber brakes, delta T = 172∘C

And the temperature change of rubber tires, delta T = 172∘C

The formula used for heat is,

Q = m s delta T

where Q is heat, m is mass, s is specific heat, and delta T is the temperature change

We know,

Heat lost by the carbon fiber brakes = Heat gained by the rubber tires

90700 g x 1.400Jg∘C x 172∘C = 123000 g x s x 172∘C

or, 90700 g x 1.400Jg∘C = 123000 g x s

or, s = (90700 g x 1.400Jg∘C) / 123000 g

or, s = 1.03236

Therefore, the specific heat of rubber tires is 1.03236 Jg∘C

8 0

3 years ago

Read 2 more answers

An alcohol is 58.80 % C and 9.889 % H by mass. The rest is oxygen. What is the empirical formula of the alcohol? Enter the eleme

PilotLPTM [1.2K]

Answer:

hco I believe

Explanation:

Pretty sure about this, when do you need to put it in?

6 0

4 years ago