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2 years ago

If a compound contains a poly atomic ion then what type of compound is it A. Convalent Compound B. Ionic Compound

Chemistry

1 answer:

den301095 [7]2 years ago

5 0

Answer:

B. Ionic Compound

Explanation:

An ionic compound is that compound which contains a positively charged ion called CATION and a negatively charged ion called ANION. The cation loses or transfers electrons to the anion, hence, making the former (cation) positive and the latter (anion) negative.

A polyatomic ion is an ion that contains more than one type of atom e.g OH-, NO3²-, CO3²- etc. A polyatomic ion usually has an overall charge formed from the charges of the individual atoms that makes it up. For example, in OH-, the overall charge is -1.

Since a polyatomic ion can have an overall positive or negative charge, it must enter a reaction with another ion that complements it i.e. a negative polyatomic ion will react with a positive ion to neutralize its charge. Hence, this forms an IONIC COMPOUND. This is why most compounds with polyatomic ions are IONIC COMPOUNDS.

For example, CaCO3 is an ionic compound formed when Ca²+ (cation) reacts with the polyatomic anion: CO3²-

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Does the moon make its own light? Explain.

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2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

4 0

3 years ago

Hydrogen gas was collected by water displacement. what was pressure of the h2 collected if the temperature was 26°c?

Vedmedyk [2.9K]

The ideal gas law may be written as
$p= \frac{\rho R T}{M}$
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)

For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol

Therefore
$p= \frac{(0.09 \, kg/m^{3})*(8.314 \, J/(mol-K))*(299 \, K)}{1.008 \times 10^{-3} \, kg/mol} =2.2195 \times 10^{5} \, Pa$

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm

Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)

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3 years ago