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IRINA_888 [86]
4 years ago
13

Write an equation to show how HC2O4− can act as a base with HS− acting as an acid.

Chemistry
1 answer:
Artyom0805 [142]4 years ago
7 0

An acid donates H^{+} ion in aqueous solution. A base accepts H^{+} ion in aqueous solution.

The equation representing the acid base reaction of HC_{2}O_{4}^{-} and HS^{-}:

HC_{2}O_{4}^{-}(aq) + HS^{-}(aq) ----> H_{2}C_{2}O_{4}(aq)+S^{2-}(aq)

In the above reaction, as HC_{2}O_{4}^{-} acts as a base it is accepting the hydrogen ion from HS^{-}. Similarly, HS^{-} donates its hydrogen ion to HC_{2}O_{4}^{-} acting as an acid.

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_______are lipids that contain cholesterol
alukav5142 [94]

Answer: Fats

Explanation: Cholesterol is one of several types of fats (lipids) that play an important role in your body.

3 0
3 years ago
How many moles are in 36.0g of H20
Nadusha1986 [10]

Answer:

The answer to your question is 2 moles

Explanation:

Data

mass of H₂O = 36 g

moles of H₂O = ?

Process

1.- Calculate the molar mass of water (H₂O)

H₂O = (1 x 2) + (16 x 1) = 2 + 16 = 18 g

2.- Use proportions and cross multiplication to find the answer.

               18 g of H₂O ---------------- 1 mol

               36 g of H₂O --------------- x

                     x = (36 x 1) / 18

                     x = 36/18

                     x = 2 moles

4 0
4 years ago
Read 2 more answers
2.122 g of a solid mixture containing only potassium carbonate (FW = 138.2058 g/mol) and potassium bicarbonate (FW = 100.1154 g/
aleksandr82 [10.1K]

Answer:

The weight percent of potassium carbonate is 50,8 wt% and of potassium bicarbonate 49,2 wt%

Explanation:

The reactions of potassium carbonate (K₂CO₃) and potassium bicarbonate (KHCO₃) with HCl produce:

K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O

KHCO₃ + HCl → KCl + CO₂ + H₂O

That means that you need 2 moles of HCl to titrate potassium carbonate and 1 mol to titrate potassium bicarbonate.

The moles of HCl to titrate the mixture are:

0,03416L×\frac{0,762mol}{1L} = <em>0,02603 mol of HCl</em>

If X is mass of K₂CO₃ and Y is mass of KHCO₃ in the mixture, the moles of HCl to titrate the mixture are equals to:

0,02603 mol = 2X×\frac{138,2058 g}{1mol} + Y×\frac{100,1154 g}{1mol} <em>(1)</em>

As the mass of the mixture is 2,122g:

2,122g = X + Y <em>(2)</em>

Replacing (2) in (1):

0,02603 mol = 0,01447 (2,122-Y) + 9,988x10⁻³Y

0,02603 mol = 0,0307 - 0,01447Y + 9,988x10⁻³Y

-4,6778x10⁻³ = -4,4827x10⁻³Y

1,044g = Y <em>-mass of potassium bicarbonate-</em>

Thus:

X = 1,078g <em>-mass of potassium carbonate-</em>

The weight percent of potassium carbonate is:

\frac{1,078g}{2,122g}×100 =<em> 50,8 wt%</em>

The weight percent of potassium bicarbonate is:

\frac{1,044g}{2,122g}×100 = <em>49,2 wt%</em>

<em></em>

I hope it helps!

5 0
3 years ago
If 5.60 mol of a substance releases 14.9 kJ when it decomposes, what is ΔH for the process in terms of kJ/mol of reactant?
lisov135 [29]

Answer:

The correct approach is "-2.67 kJ/mole". A further solution is described below.

Explanation:

The given values are:

No. of moles,

= 5.60 mol

Substance releases,

ΔH = 14.9 kJ

Now,

The ΔH in terms of kJ/mol will be:

= \frac{\Delta H}{5.60}

On putting the given values of ΔH, we get

= \frac{-14.9}{5.60}

= -2.67 \ kJ/mole

8 0
3 years ago
An airplane weighing 80,000. kg at takeoff reaches a speed of 880 km/h. What is the de Broglie wavelength in meters of the plane
Rzqust [24]

Answer:

3.39 * 10^-41 m

Explanation:

Given that:

speed of airplane = 880 km/hr

Mass of airplane = 80000 kg

Debroglie wavelength (λ) = h / mv

h = planck's constant = 6.626 * 10^-34

; mass = 80000kg

Velocity = 880km/hr to convert to m/s

Velocity = (880 * 1000m) / 3600 =244.44 m / s

λ = (6.626 * 10^-34) /(80000*244.44)

λ = (6.626 * 10^-34) / 19555200

λ = 3.3889 *10^-41

λ = 3.39 * 10^-41 m

6 0
3 years ago
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