1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Black_prince [1.1K]
3 years ago
8

6th grade science i mark as brainliest​

Chemistry
2 answers:
MaRussiya [10]3 years ago
7 0
C is the correct answer
valentinak56 [21]3 years ago
5 0

Answer:

the answer you have marked is correct.

C is correct.

You might be interested in
The following equation represents the type of reaction called
Mazyrski [523]

Answer:

Reduction

Explanation:

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Mn⁺⁷ +3e⁻    →    Mn⁴⁺

Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.

Examples:

Consider the following reactions.

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

6 0
2 years ago
Given the following thermodynamic data, calculate the lattice energy of LiCl:
tiny-mole [99]

Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

3 0
2 years ago
Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50
devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C_{V} You can work out C_{V}

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC_{V} ∆T

Polyatomic gas: C_{V} = 3R

∆U= nC_{V} ∆T

∆U= 28g x C_{V} x (350K - 58.3K)

∆U = 28C_{V} x 291.7

∆U = 10967.6 x C_{V}

5 0
3 years ago
Rank the following elements by electron affinity, from most positive to most negative EA value.
LenKa [72]

Answer:

Explanation:

Bromine >Tellurium > Phosphorus > Helium > Sodium

Electron affinity of Bromine , Tellurium , Phosphorus are positive , of helium is zero and of sodium is negative .

3 0
3 years ago
WILL MARK BRAINLIEST. QUICK
Murrr4er [49]

The Answer is D: Reactor products charge huge batteries.

6 0
2 years ago
Read 2 more answers
Other questions:
  • What does not exist in a supersaturated solution?
    14·1 answer
  • HELP 30 Points !
    5·2 answers
  • Which of the following liquids will have the highest surface tension?
    11·1 answer
  • Triphenylmethane can be prepared by reaction of benzene and chloroform in the presence of AlCl3. Draw curved arrows to show the
    14·1 answer
  • If a sodium atom lost one of its electrons and a chlorine atom gained it, what kind of bond would be formed between sodium and c
    10·2 answers
  • Susan needs to simplify the expression shown below what should be her first step
    10·1 answer
  • When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)
    8·1 answer
  • PLZ HELP!!!!!!!!!!!!!!!!!!!!!!!!!Describe Earth's rotation and revolution . Be sure to include the durations and what they are r
    13·1 answer
  • Summarize our current
    12·1 answer
  • In this project, you will practice calculating work and speed using the proper formula. You will also use a data table to record
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!