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
Kruka [31]
3 years ago
13

Match the element with its number of valence electrons. Column A 1. Boron : Boron 2. Carbon : Carbon 3. Fluorine : Fluorine 4. S

odium : Sodium Column B a. 4 b. 1 c. 3 d. 7
Chemistry
1 answer:
Paha777 [63]3 years ago
6 0
Boron 3
Carbon 4
Fluorine 7
Sodium 1
You might be interested in
150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac
Murljashka [212]

Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

6 0
3 years ago
A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
6 0
3 years ago
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
100 POINTS PLEASE HELP (PHOTOS INCLUDED)
pshichka [43]

Answer:

Explanation:

3.

Knowns: 100mL of solution; concentration of 0.7M

Unknown: number of moles

Equation: number of moles = volume * concentration

Plug and Chug: number of moles = 100/1000 * 0.7 = 0.07 mole

Final Answer: 0.07mole

2.

Knowns: 5.50L of solution; concentration of 0.400M

Unknown: number of moles

Equation: number of moles = volume * concentration

Plug and Chug: number of moles = 5.5 * 0.4 = 2.20 mole

Final Answer: 2.20 mole

6 0
2 years ago
Read 2 more answers
When the heat of reaction (H=ve), the reaction is Endothermic reaction<br> True<br> False
juin [17]
False because that doesn’t make sense
6 0
3 years ago
Read 2 more answers
Other questions:
  • How do you calculate the number of neutrons in atom
    5·1 answer
  • What is the emf produced in a 1.5 meter wire moving at a speed of 6.2 meters/second perpendicular to a magnetic field of strengt
    5·2 answers
  • When potassium carbonate and calcium chromate are mixed, which of the following will pass through a filter?
    15·1 answer
  • Please answer ASAP!!!
    12·1 answer
  • Wait is there a difference between number of mol and number of moles? Or do they mean the same thing lol
    14·1 answer
  • Compare and contrast the three states of matter liquid, solid, and gas.
    15·1 answer
  • What happens when an electrically charged pencil is placed close to the water running from a faucet?
    13·2 answers
  • BRAINLIEST FOR THE FIRST CORRECT ANSWER!! im revising for a test and my teacher gave me some questions to do but i dont understa
    5·1 answer
  • Is Fe2+ + 2e- = Fe an oxidation or reduction
    14·1 answer
  • Transcribe and translate the following DNA sequence:
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!