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3 years ago

In the reaction A+B = C+D if B is the excess what is the limiting reactant

Chemistry

1 answer:

xenn [34]3 years ago

7 0

Explanation:

Determining the Limiting Reagent

One reactant (A) is chosen, and the balanced chemical equation is used to determine the amount of the other reactant (B) necessary to react with A. If the amount of B actually present exceeds the amount required, then B is in excess, and A is the limiting reagent.

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Will give brainlest if correct

cupoosta [38]

Answer:

igneos

Explanation:

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2 years ago

Read 2 more answers

Iridium has only two naturally occurring isotopes. The mass of iridium-191 is 190.9605 amu and the mass of iridium-193 is 192.96

cluponka [151]

Answer:

Abundance of Iridium-193 is 62.75%

Explanation:

From the question given above, the following data were obtained:

Isotope A (Iridium-191):

Mass of A = 190.9605 amu

Abundance of A = A%

Isotope B (Iridium-193):

Mass of B = 192.9629 amu

Abundance B = (100 – A) %

Relative atomic mass of Iridium = 192.217 amu

Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:

Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]

192.217 = 1.909605A% + 1.929629(100 – A)%

192.217 = 1.909605A% + 192.9629 – 1.929629A%

Collect like terms

192.217 – 192.9629 = 1.909605A% – 1.929629A%

–0.7459 = –0.020024A%

Divide both side by –0.020024

A% = –0.7459 / –0.020024

A% = 37.25 %

Finally, we shall determine the abundance of Isotope B (Iridium-193).

This can be obtained as follow:

Abundance of A (Iridium-191) = 37.25 %

Abundance of B (Iridium-193) =?

Abundance B = 100 – A%

Abundance B = 100 – 37.25 %

Abundance of B (Iridium-193) = 62.75%

4 0

2 years ago

If kb for nx3 is 4.0×10−6, what is the poh of a 0.175 m aqueous solution of nx3?

Elena-2011 [213]

<h3><u>Answer;</u></h3>

pOH = 3.08

<h3><u>Explanation;</u></h3>

NX3 + H2O <----> NHX3+ + OH-

Kb = 4.0 x 10^-6

Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).

c(NH₄⁺) = c(OH⁻) = x.

x² = Kb · c(NH₃)

x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = c(OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

7 0

3 years ago

Which of the following elements is most likely to form a covalent bond?

Alecsey [184]

Bromine, it's the only element that isn't a metal

7 0

3 years ago

What is the percent-by-mass concentration of acetic acid (CH3COOH) in a vinegar solution that contains 51.80 g acetic acid in a

Semenov [28]

Answer : The percent-by-mass concentration of acetic acid is, 5.15 %

Explanation : Given,

Mass of acetic acid = 51.80 g

Density of solution = 1.005 g/ml

Volume of solution = 1.000 L = 1000 ml (1 L = 1000 ml)

First we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}$

$\text{Mass of solution}=1.005g/ml\times 1000ml=1005g$

Now we have to calculate the percent-by-mass concentration of acetic acid.

$\% \text{ by mass of acetic acid}=\frac{\text{Mass of acetic acid}}{\text{Mass of solution}}\times 100$

Now put all the given values in this formula, we get:

$\% \text{ by mass of acetic acid}=\frac{51.80g}{1005g}\times 100=5.15\%$

Therefore, the percent-by-mass concentration of acetic acid is, 5.15 %

6 0

2 years ago