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3 years ago

Which is a spectator ion from the following complete ionic equation:

2 answers:

8 0

K+(aq) and Br-(aq)

Because the reactant also appears on the product side in the same form, it does not go through any chemical reaction and is therefore a spectator ion.

kompoz [17]3 years ago

7 0

Answer : The spectator ions are, $K^+\text{ and }Br^-$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given ionic equation in separated aqueous solution will be,

$H^+(aq)+Br^-(aq)+K^+(aq)+OH^-(aq)\rightarrow H_2O(l)+K^+(aq)+Br^-(aq)$

In this equation, $K^+\text{ and }Br^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$H^+(aq)+OH^-(aq)\rightarrow H_2O(l)$

Hence, the spectator ions are, $K^+\text{ and }Br^-$

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Name one alternative pathway to glycolysis: _______________________

NemiM [27]

solution:

The Enter–Doudoroff pathway describes an alternate series of reactions that catabolize glucose to pyruvate using a set of enzymes different from those used in either glycolysis or the pentose phosphate pathway. .... Most bacteria use glycolysis and the pentose phosphate pathway.

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4 years ago

In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00kg o

Lerok [7]

M = amount of the solute / mass of the solvent

0.523 = x / 2.00

x = 0.523 * 2.00

x = 1,046 moles

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Mass = 1,046 * 166.0028

Mass ≈ 173.63 g

hope this helps!

6 0

4 years ago

E85 fuel is a mixture of 85% ethanol and 15% gasoline by volume. What mass in kilograms of E85 (d = 0.758 g/mL) can be contained

NISA [10]

Convert gal to liters:

14.0 gal x 3.785 L / gal = 53.0 L

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4 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

Answer: The limiting reactant is ammonia $(NH_3)$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For ammonia:

Given mass of ammonia = 135.9 kg = 135900 g (Conversion factor: 1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol$

For carbon dioxide gas:

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol$

The given chemical reaction follows:

$2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = $\frac{1}{2}\times 7994.12=3997.06mol$ of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia $(NH_3)$

5 0

3 years ago

How many Sig Figs these numbers contain ? 1-2.008 2- 0.0253 3-15000

Dmitry_Shevchenko [17]

1) has 4 sig figs(0s between 2 significant digits are important)
2) has 3 sig figs (the trailing 0s aren’t significant)
3) has 2 sig figs (the trailing 0s aren’t significant)

5 0

4 years ago