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3 years ago

Identify the spectator ions in the following complete ionic equation.

1 answer:

5 0

Ba2+(aq)+2I−(aq)+2Na+(aq)+SO2−4(aq)→BaSO4(s)+2I−(aq)+2Na+(aq)

The spectator ions are: Na⁺, I⁻.

They appear on both sides of the reaction, hence did not take part in the reaction.

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Nitrogen and hydrogen gases are combined at high temperatures and pressures to produce ammonia, NH3. If 101.7 g of N2 are reacte

Lerok [7]

Answer:

7.26 moles of NH₃ are formed in this reaction

Explanation:

This is about the reaction for the production of ammonia

1 mol of nitrogen gas reacts to 3 moles of hydrogen in order to produce 2 moles of ammonia.

The equation is: N₂ + 3H₂ → 2NH₃

In the question, we were informed that the excess is the H₂ so the N₂ is limiting reagent. We determine the moles, that has reacted:

101.7 g / 28 g/mol = 3.63 moles

So, If 1 mol of nitrogen gas can produce 2 moles of ammonia

3.63 moles of N₂ must produce ( 2 . 3.63) / 1 = 7.26 moles of NH₃

6 0

4 years ago

Read 2 more answers

___C + ___S8 → ___CS2

qwelly [4]

Answer:

4,1,4

Explanation:

Balancing a synthesis equation

6 0

3 years ago

Sodium (Na) reacts with chlorine gas (Cl2) to form sodium chloride (NaCl) as shown in the equation 2Na+Cl-->2NaCl. If 4 moles

-Dominant- [34]

For every 2 mol of Na 1 mole of cl2 will react
4 mol Na = 2mole cl2

8 0

3 years ago

When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s)+CO2(

Mazyrski [523]

Answer is: mass of calcium carbonate needed is 120 grams.
Chemical reaction: CaCO₃(s) → CaO(s) + CO₂(g).
V(CO₂) = 27.0 L.
Vm = 22.4 L/mol.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 27 L ÷ 22.4 L/mol.
n(CO₂) = 1.2 mol.
From chemical reaction: n(CO₂) : n(CaCO₃) = 1 : 1.
m(CaCO₃) = 1.2 mol.
m(CaCO₃) = n(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 1.2 mol · 100 g/mol.
m(CaCO₃) = 120 g.

3 0

4 years ago

PLEASE HELP FAST IM ABOUT TO FAIL PLEASE HELP PLEASE HELP FAST HELP HELP THIS IS DESPRATE

MariettaO [177]

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= $\frac{mass}{atomic mass of one mole}$ 1st equation

molarity = $\frac{number of moles}{volume}$ 2nd equation

Dilution formula

M1V1 = M2V2 3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = $\frac{21.6}{68.946}$

= 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = $\frac{0.313}{1.3}$

= 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = $\frac{215.1}{36.46}$

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = $\frac{5.899}{2}$

= 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

M1= 12 M

V2=?

M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119 moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = $\frac{0.0119}{0.1}$

= 0.11 M (pOH Concentration)

14 = pH + pOH

pH = 14 - 0.11

pH = 13.89

3 0

3 years ago