I can be formed into a sheet, I am high in luster, and am a very good conductor. I must be a ...

Enter a balanced equation for the reaction between aqueous lead(II)(II) nitrate and aqueous sodium chloride to form solid lead(I

aev [14]

Answer:

Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) ↓ + 2NaNO₃ (aq)

Explanation:

The reactants are:

Lead(II) nitrate → Pb(NO₃)₂ (aq)

Sodium chloride → NaCl (aq)

The products are:

Lead(II) chloride → PbCl₂ (s)

Sodium nitrate → NaNO₃ (aq)

Salts form nitrate are soluble. The chloride makes a precipitate with the Pb²⁺. The chemical equation for this reaction is:

Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) ↓ + 2NaNO₃ (aq)

3 0

4 years ago

All circuits must have three things: energy source, wire, and a generator Lesson 5.04 Question 7 options: True False

Doss [256]

Answer:

True

Explanation:

Sorry i'm late probably but i answered this for you :) i hope this helps though

5 0

3 years ago

An ice skater moves back and forth on an ice rink. She speeds up on the straight portion of the rink. She slows down near each e

andrey2020 [161]

The options are;

A) The skater experiences acceleration both while skating in a straight line and while making her turn.

B) Because the skater does not change her direction or speed, she does not experience acceleration.

C) The skater only experiences acceleration when she is turning.

D) The skater only experiences acceleration while skating in a straight line.

Answer:

A: The skater experiences acceleration both while skating in a straight line and while making her turn

Explanation:

We are told that She speeds up on the straight portion of the rink. This means that she experienced an acceleration since the speed was not constant but gradually increasing.

Also, we are told that She slows down near each end of the rink and then turns around. This means that she reduced speed before turning and it means there was also an acceleration as well when making the turn.

Thus, she experience acceleration both in the straight line and when making turn.

Option A is correct

6 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

4 years ago

Use this condensed chemical structure

liberstina [14]

Answer:

number of carbon-carbon single (C - C) bonds: 1

number of carbon-hydrogen single (C H) bonds: 5

number of nitrogen-hydrogen sing le (N H) bonds:2

number of lone pairs: 1

Explanation:

Ethanamine is a colourless gas having a strong 'ammonia- like' odour. It contains the -NH2 group which makes it an amine. It contains one carbon-carbon bond, five carbon-hydrogen bonds and two nitrogen-hydrogen bonds.

Nitrogen, being sp3 hybridized in the compound has a lone pair of electrons localized on one of the sp3 hybridized orbitals of nitrogen while one sp3 hybridized orbital of nitrogen is used to form a carbon-nitrogen bond. The other two sp3 hybridized orbitals on nitrogen are used to form the two nitrogen-hydrogen bonds.

6 0

3 years ago