All categories

3 years ago

Why can’t you perform single displacement reactions with group IA metals?

1 answer:

5 0

Answer is: because alkaline metals (group IA metals) are the strongest reducing agents and most reactive metals.
Reducing agent is an element or compound that loses an electron to another chemical species in a redox chemical reaction and they have been oxidized.
Alkaline metals tend to lose only one electron in redox reaction.

You might be interested in

The form of EMR that has less energy per photon than microwaves is ____?

Gala2k [10]

Answer: radio wave frequencies have longer wavelengths and smaller

Energies per photon

4 0

3 years ago

Suppose 2.90g of nickel(II) iodide is dissolved in 150ml of a 0.70M aqueous solution of potassium carbonate. Calculate the final

ehidna [41]

Answer:

0.124 M.

Explanation:

Hello!

In this case, since the nickel iodide has the following formula:

NiI₂

So its molar mass is 312.5023 g/mol, in order to compute the molarity of the iodide anion, we first need the moles in 2.90 g:

$n_{NiI_2}=2.90g*\frac{1molNiI_2}{312.5023 gNiI_2} =0.00927molNiI_2$

Now, since one mole of nickel(II) iodide contains two mole of iodide anions, we infer there are 0.0186 moles of iodide cations. Moreover, since the molarity is computed by dividing the moles of those ones by the volume of the solution in liters, 150 mL (0.150 L) as it does not change, it turns out:

$M=\frac{0.0186molI^-}{0.150L}\\\\M=0.124M$

Best regards!

8 0

3 years ago

What is the empirical formula for glucose A) B) C) D)

pshichka [43]

The Answer is A CH2O

The molecular formula for glucose is C6H12O6. The subscripts represent a multiple of an empirical formula. To determine the empirical formula, divide the subscripts by the GCF of 6 which gives CH2O.

4 0

3 years ago

Read 2 more answers

What is the oxidation state of sulfur in SO3?

alex41 [277]

The oxidation state of a substance is the electric charge it is exhibiting in a given state. This may be determined by looking at the oxidation states of accompanying atoms as well as the charge on the complete molecule.

In this case:

Molecular charge: 0
Oxidation state of oxygen: -2 (it is a group 6 element)

Thus,

S + 3 * -2 = 0
S = 6

Sulfur is exhibiting an oxidation state of +6.

8 0

3 years ago

Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation 4 P+5O

icang [17]

Answer:

7.03g

Explanation:

4P + 5O2 → 2P2O5

Let us convert the mass given to mole. This can be achieved by doing the following:

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 = 6.09g

Number of mole = Mass /Molar Mass

Number of mole of O2 = 6.09/32 = 0.19mol

Molar Mass of P = 31g/mol

Mass of P = 3.07g

Number of mole of P = 3.07/31 = 0.099mol

Let us determine the limiting reactant and the excess reactant

From the equation,

4moles of P required 5moles of O2.

Therefore, 0.099mol of P will require = (0.099 x 5)/4 = 0.12mol

From the above illustration, we see clearly that not all the O2 reacted as the number of mole of O2 obtained from the question is 0.19mol. This means that O2 is the excess reactant and P is the limiting reactant.

Note: the limiting reactant is always used to obtain the yield of any reaction.

Now we can obtain the theoretical yield of P2O5 as follows:

4P + 5O2 → 2P2O5

Molar Mass of P = 31g/mol

Mass of P from the equation = 4x31 = 124g

Molar Mass of P2O5 = (31x2) + (16x5) = 62 + 80 = 142g/mol

Mass of P2O5 from the equation = 2 x 142 = 284g

From the equation,

124g of P produced 284g of P2O5.

Therefore, 3.07g of P will produce = (3.07x284)/124 = 7.03g of P2O5.

Therefore, the theoretical yield of P2O5 is 7.03g

4 0

4 years ago