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3 years ago

11

What is produced during the replacement reaction of Cu(NO3)2 and Zn?

1 answer:

PolarNik [594]3 years ago

8 0

During the displacement reaction , copper (cu) and zinc nitrate (<span> 2 ZnNo<span>3 )</span></span> are the products . hope this helps :)

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All atoms of an element are ________?

mote1985 [20]

Possibly different But here I hope this helps Recall that neutrons, along with protons, make up the nucleus. Since neutrons have no electrical charge, they do not affect the chemical behavior. ... Atoms of the same element with different numbers of neutrons are called isotopes.

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2 years ago

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The mass of a jar of sugar is 11. 9 kg. What is the total mass of 13 jars of sugar?.

Masteriza [31]

Answer:

154.7

Explanation:

11.9kg x 13 = 154.7

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2 years ago

Classify the following reactions into: Element – Element, Element – Compound, Compound – Compound

Rudiy27

Answer:

1. CaO + H₂O ----> Ca(OH)₂

Compound ----- Compound

2. 2 Na + Cl₂ ----> 2 NaCl

Element ----- Element

3. 2 SO₂ + O₂ ----> 2 SO₃

Element ----- Compound

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1 year ago

WILL MARK BRAINLIEST

den301095 [7]

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2 years ago

20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator

Eduardwww [97]

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to 5.6

Explanation:

The equation for the reaction is :

$C_2H_5NH_2_(_a_q_) + H^+_(_a_q_) --- C_2H_5NH_{3(aq)}^+$

concentration of $C_2H_5NH_{2(aq)$ = 10%

10 g of $C_2H_5NH_{2(aq)$ in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of $C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL$

= 2.22 M

number of moles of $C_2H_5NH_{2(aq)$ in 20 mL can be determined as:

$= 20 mL \times 2.22 M= 44*10^{-3} mole$

Concentration of $C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}$

= 2.22 M

Similarly, The pKa Value of $C_2H_5NH_{2(aq)$ is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

$pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]$

$pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]$

$pH = 5.21$

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

8 0

3 years ago