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3 years ago

Which is an anion ca2+ ag+ o2- k+

Chemistry

2 answers:

igor_vitrenko [27]3 years ago

8 0

Answer:

O2- is an anion

Explanation:

An atom or molecule is said to be neutral when it has a zero net charge. In contrast, if the system has a net charge it is termed as an ion.

A cation is an atom or a molecular system with a net positive charge whereas an anion carries a net negative charge.

In the given examples:

Ca2+ : net positive, cation

Ag+: net positive, cation

O2-: net negative, anion

K+: net positive, cation

a_sh-v [17]3 years ago

6 0

The 3rd one.
O2-
It is an Anion because the Oxygen atom has one extra electron from another atom.

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What volume of 0.160 m li2s solution is required to completely react with 130 ml of 0.160 m co no3 2?

Over [174]

The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below

write the reacting equation

Co(NO3)2 + Li2S = 2LiNO3 + COS

find the moles of CO(NO3)2 = molarity x volume

= 130 ml x 0.160=20.8 moles

since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles

volume of Lis is therefore = moles of Lis/ molarity of LiS

= 20.8/0.160 = 130 Ml

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3 years ago

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

6 0

3 years ago

A large cyclotron directs a beam of He++ nuclei onto a target with a beam current of 0.250 mA. (a) How many He++ nuclei per seco

nikitadnepr [17]

Answer:

a. 7.8*10¹⁴ He⁺⁺ nuclei/s

b. 4000s

c. 7.7*10⁸s

Explanation:

I = 0.250mA = 2.5 * 10⁻³A

Q = 1.0C

1 e- contains 1.60 * 10⁻¹⁹C

But He⁺⁺ Carrie's 2 charge = 2 * 1.60*10⁻¹⁹C = 3.20*10⁻¹⁹C

(A).

No. Of charge per second = current passing through / charge

1 He⁺⁺ = 2.50 * 10⁻⁴ / 3.2*10⁻¹⁹C

1 He⁺⁺ = 7.8 * 10¹⁴ He⁺⁺ nuclei

(B).

I = Q / t

From this equation, we can determine the time it takes to transfer 1.0C

I = 1.0 / 2.5*10⁻⁴ = 4000s

(C).

Time it takes for 1 mol of He⁺⁺ to strike the target =?

Using Avogadro's ratio,

1.0 mole of He = (6.02 * 10²³ ions/mol ) * (1 / 7.81*10¹⁴ He ions)

Note : ions cancel out leaving the value of the answer in mols.

1.0 mol of He = 7.7 * 10⁸s

8 0

3 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$