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3 years ago

What is the concentration that of chloride ions in a .193 m solution of potassium chloride

Chemistry

1 answer:

xenn [34]3 years ago

5 0

Answer:

solution of K3PO4?

1.31 m

Concentration of Chloride Ions in a 0.193m solution of Potassium Chloride is _______.

0.193 m

Concentration of Iodide ions in a 0.193 solution of Barium iodide is _________.

0.386 m

In a Titration of 35.00 mL of 0.737 m H2So4, _______ mL of a 0.827 m KOH solution is required for neautralization.

62.4

Consider the following reactions

AgNo3 ------>

Co(No3) ---->

AgNo3 ---->

Which is the correct order of increasing activity for these metals?

Ag < Co < Zn

There are ______ Mol of bromide ions in 0.500 L of a 0.300M solution of AlBr3

0.450 m

How many grams of Sodium Chloride are there in 550 mL of a 1.90 m aq solution of sodium CL?

6.11 g

The total concentration of ions in a 0.250 M solution of HCL is _______?

0.500 m

How many moles of Co 2+ are present in 0.200 L of a 0.400 M solution of Co I2

0.0800

10.

A Neautralization reaction between an acid and a metal hydroxide produces

Water and a salt

11.

7 Strong Acids

Claire HCl

Brian HBr

invented HI

nutrious HNo3

(3)candy HClO3

(2,4)sold H2So4

(4)profit HClO4

12.

8 strong bases

Lily LiOH Lithium Hydroxide

never NaOH Sodium Hydroxide

knew KOH PossasiumHydroxide

Robert RbOH

could ClOH

bake Ba(OH)2

strawberry Sr(OH)2

cake Ca(OH)2

13.

What are the spectators ions in the reaction between KOH(aq) and HNO(aq)

HF + OH ---> H2O + F-

14.

Combining Aq solution of BaI2 and Na2So4 affords a precipitate of BaSO4

Which Ions is/are spectator ions in the reaction?

Na + and I -

15.

The Spectator ions in the react between aq Hydrochloric acid and Aq Ammonia

Are ________

B. Cl - only

16.

Spectator ions of aq hydrofluoric acid and aq barium hydroxide are _________

D. Ba 2+ only

17.

Spectator ions in reactions

Aq perchloric acid and Aq barium hydroxide are >>>>>>>

ClO 4- and Ba 2+

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If an ice cube weighing 25.0 g with an initial

riadik2000 [5.3K]

Answer:

∘

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

∘

to liquid at

∘

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

−

(

)

, where

- the heat absorbed by the solid at

∘

- the heat absorbed by the liquid at

∘

- the heat lost by the warmer water sample

The two equations that you will use are

⋅

, where

- heat absorbed/lost

- the mass of the sample

- the specific heat of water, equal to

4.18

∘

- the change in temperature, defined as final temperature minus initial temperature

and

⋅

fus

, where

- heat absorbed

- the number of moles of water

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

⋅

1 mole H

18.015

5.551 moles H

So, how much heat is needed to allow the sample to go from solid at

∘

to liquid at

∘

5.551

moles

⋅

6.01

mole

33.36 kJ

This means that equation

(

)

becomes

33.36 kJ

−

The minus sign for

is used because heat lost carries a negative sign.

So, if

is the final temperature of the water, you can say that

33.36 kJ

sample

⋅

sample

−

water

⋅

water

More specifically, you have

33.36 kJ

100.0

⋅

4.18

∘

⋅

(

−

)

∘

−

650

⋅

4.18

∘

⋅

(

−

)

∘

33.36 kJ

418 J

⋅

(

−

)

−

2717 J

⋅

(

−

)

Convert the joules to kilojoules to get

33.36

0.418

⋅

−

2.717

⋅

(

−

)

This is equivalent to

0.418

⋅

2.717

⋅

67.925

−

33.36

34.565

0.418

2.717

11.026

∘

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

∘

Explanation:

3 0

3 years ago

For the chemical reaction , I identify the reactant and the products .

svetlana [45]

The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.

This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.

Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.

Hope this helps!

7 0

3 years ago

For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of

viktelen [127]

Answer:

$\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%$

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

$n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\$

Now, we compute the atomic percentages as shown below:

$\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%$

Best regards!

4 0

3 years ago

Explain how condensation is invovled in the water cycle

blagie [28]

Condensation is the process of watervapor in the air is changing into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds. Condensation is also the opposite of evaporation

8 0

3 years ago

In your own words list the rules for naming hydrocarbons and substituted hydrocarbons. Be detailed in listing all rules includin

Irina18 [472]

Answer :

The basic rules for naming of hydrocarbons are :

First select the longest possible carbon chain.

The longest possible carbon chain should include the carbons of double or triple bonds.

The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne.

The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.

The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.

If two or more similar alkyl groups are present in a compound, the prefixes di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

8 0

4 years ago