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2 months ago

The dihydrogenphosphate ion, H2PO4− is amphiprotic. In which of the following reactions is this ion serving as a base?

Chemistry

1 answer:

Anarel [2.9K]2 months ago

4 0

Answer:

H3PO4(aq) + HPO42-(aq) ⇄ 2H2PO4-

Explanation:

During the process involving Phosphonic acid, H3PO4, a proton H+ is eliminated, resulting in the production of the dihydrogen phosphate ion, H2PO4-, which acts as a base.

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A syringe contains 0.65 moles of he gas that occupy 900.0 ml. what volume (in l) of gas will the syringe hold if 0.35 moles of n

lions [2927]

<span>Using PV=nRT, which represents a universal constant for any state, we have: P1V1/n1T1=R and P2V2/n2T2=R; This implies that: P1V1/n1T1=P2V2/n2T2 Thus we can express it as V1/n1=V2/n2. Rearranging yields: V2=V1 x (n2/n1) = 750 mL x ((0.65+0.35)/(0.65)) = 1200 mL = 1.2 L... with 2 significant figures</span>

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2 months ago

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Which element has the electron configuration [Xe] 6s2 4f14 5d10 6p2?

eduard [2782]

The element you are looking for is Pb (Lead). Just check the last orbital on the periodic table to find it!

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1 month ago

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5 grams of NaHCO3 was added to 50 cm3 of 1.5 mol dm-3 hydrochloric acid, causing the temperature to decrease by 6.8 K. What is t

Anarel [2989]

The balanced chemical equation for the neutralization of HCl with $NaHCO_{3}$ is:

$HCl (aq) + NaHCO_{3}(aq)--> NaCl (aq) + H_{2}O(l)+CO_{2} (g)$

Given weight of $NaHCO_{3}$ = 5g

Moles of $NaHCO_{3}$ = $5 g *\frac{1 mol}{84 g} = 0.05952 mol NaHCO_{3}$

Volume of HCl solution = $50 cm^{3} * \frac{1 mL}{1cm^{3}} = 50 mL$

Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184 $\frac{J}{g. ^{0}C}$

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2 $^{0}C$

$Q = 55 g * 4.184 \frac{J}{g. K}(6.8K) = 1565 J$

The enthalpy of neutralization per mole of $NaHCO_{3}$

= $\frac{1565J}{0.05952 mol} = 26294 \frac{J}{mol}*\frac{1 kJ}{1000J} =26.294kJ/mol$

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3 months ago

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If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr

VMariaS [2998]

Answer:

The heat capacity of the calorimeter is $C_c$ = 54.4 $\frac{J}{c}$

Explanation:

Given the data

Heat supplied Q = 4.168 KJ = 4168 J

Mass of water $m_w$ = 75.40 gm

Change in temperature = ΔT = 35.82 - 24.58 = 11.24 °C

From the conditions provided

Q = $m_w C_w$ ΔT + $C_c$ ΔT

Plugging all values into the above equation yields

4168 = 75.70 × 4.18 × 11.24 + $C_c$ × 11.24

611.37 = $C_c$ × 11.24

$C_c$ = 54.4 $\frac{J}{c}$

This represents the heat capacity of the calorimeter.

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2 months ago

In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form

KiRa [2933]

I predict that there will be an increase in the seconds recorded in the time column. This is because, as more water is mixed with sodium thiosulfate, its concentration diminishes in each flask. Additionally, a lower concentration results in a slower reaction rate since fewer molecules of sodium thiosulfate means there are less frequent collisions with sulfuric acid. With fewer collisions occurring in the reaction, it takes a longer time for the reaction to complete, leading to increased time when sodium thiosulfate is diluted.

Explanation:

I can confirm that this explanation is accurate.

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3 months ago