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4 years ago

Which has a higher melting point, NaBr or KBr

Chemistry

2 answers:

Blababa [14]4 years ago

7 0

NaBr melting point : 1,377°F (774C)
KBr: 1,353°F (734C)

Oduvanchick [21]4 years ago

6 0

Answer:

NaBr

Explanation:

Both compounds are electrovalent compounds, that is they are held together by electrovalent combination. Potassium bromide (KBr) has a melting point of 734 °C while sodium bromide (NaBr) has a melting point of 747 °C. The reason for this is because the electropositivity of potassium is higher than that of sodium.

Electrovalent combination involves the transfer the electrons from one constituent atom (to form a cation) to another constituent atom (to form an anion). Electropositivity is the tendency of an atom to donate electron(s) to form a cation. Since, it's easier for potassium (when compared to sodium) to donate its outermost electron to bromine, NaBr will have a higher force of attraction than KBr and hence a higher melting point.

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3 years ago

URGENT! A solution that contains less solute than it can possibly hold at a given temperature.

Zigmanuir [339]

Answer:

Diluted solution

Explanation:

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4 years ago

Anhydrous calcium chloride removes water vapor from the air inside a container having an air-tight seal called a desiccator. Exp

OlgaM077 [116]

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3 years ago

Which of the following are true of phases?Check all that apply.A.The particles in a liquid are in a fixed location.B.Gases have

saw5 [17]

Answer:

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7 0

3 years ago

Complete and balance the following redox reaction in acidic solution As2O3(s) + NO3- (aq) → H3AsO4(aq) + N2O3(aq)

scoray [572]

Answer:

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Explanation:

Oxidation: $As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)$

Balance As: $As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)$
Balance H and O in acidic medium: $As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$
Balnce charge: $As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$ ......(1)

Reduction: $NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$

Balance N: $2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$
Balance H and O in acidic medium: $2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$
Balance charge: $2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$ ......(2)

$Equation (1)+Equation (2)$ gives-

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

3 0

3 years ago

Read 2 more answers