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nevsk [136]
3 years ago
14

Which of the following tends to form anions when bonding with other elements?

Chemistry
2 answers:
r-ruslan [8.4K]3 years ago
8 0

Answer:

ca

Explanation:

i got it right on the test

yarga [219]3 years ago
3 0

Answer: it’s D

Explanation:

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Water, lower, and I'm not too sure for the third one... sorry.
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A) For the following reaction, K > 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry a
Aleks04 [339]

Answer:

Explanation:

According to Bronsted-Lowry acids or base theory ,  the reagent capable of giving hydrogen ion or proton  will be acid and that which accepts hydrogen ion or proton  will be base .

C₉H₇N + HNO₂   ⇄    C₉H₇NH⁺ + NO₂⁻

If K > 1 , reaction is proceeding from left to right .

Hence HNO₂ is giving H⁺ or proton and C₉H₇N is accepting proton to form

C₉H₇NH⁺ .  

Hence HNO₂ is bronsted acid and C₉H₇N is bronsted base .

B )

when K < 1 , reaction above proceeds from right to left . That means

C₉H₇NH⁺ is giving H⁺ so it is a bronsted acid and NO₂⁻ is accepting H⁺ so it is a bronsted base .

Hence ,  NO₂⁻ is a bronsted base and C₉H₇NH⁺ is a bronsted acid .

4 0
3 years ago
What is the ph of a saturated solution of aluminum hydroxide?
sasho [114]
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6 0
3 years ago
How can we find the volume of this fish tank? Please I needed but Good
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3 0
3 years ago
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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
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