5.6 g of an organic compound on burning with excess
of oxygen gave 17.6 g of CO2, and 7.2 g of H,O. The
organic compound is :​

Answer:

10 the power of 23 hydrogen atom

Answer:

Explanation:

One farad of charge is capable of depositing one gram equivalent of a metal

One gram equivalent of Ag = 108 grams

One gram equivalent of Cd = 112 / 2  grams

= 56 grams . [ for cadmium equivalent mass = atomic mass / s ]

electric charge flowing = current x time = 5 x 600 = 3000 coulomb

one farad = 96500 coulomb

96500 coulomb  deposits 108 gram of Ag

3000 coulomb deposits 108 x 3000 / 96500 gram

= 3.35 grams of Ag

Similarly ,

96500 coulomb  deposits 56 gram of Cd

3000 coulomb deposits 56 x 3000 / 96500 gram

= 1.74  grams of Cd

So there will be much larger increase in the spoon of Ag due to larger deposit of Ag by charge .

Answer:

i think its Yes. The bubbles contain hydrogen and oxygen that separated from the water but i could be wrong so i would just wait to see if anybody else says anything

Explanation:

Answer:

The sand bath spreads the heat out so that the flask is heated evenly. This reduces the chance of the flask breaking and ensures that there are no hot spots in the reaction mixture which could lead to excessive charring,

Explanation:

Answer:

Yes

Yes

Yes

All yes

Explanation:

Answer:

9th:08473 kJ/mol

Explanation:

HAHAHAHAHAHAHAHAHAHA

Answer:

true

Explanation:

you are getting closer to the core hope this helps!

Answer:

1. A

2. B

3. B

4. A

5. B

Explanation:

1. Recommended. That way cross-contamination is prevented.

2. Not recommended. The inner mechanism of the micropipette should not come in contact with any liquid whatsoever.

3. Not recommended. It contradicts statement 1.

4. Recommended. That way the micropipette remains completely vertical, minimizing the risks of being decalibrated.

5. Not recommended. If it lays on the bench it will be horizontal, thus contradicting statement 4.

Answer:

1. A

2. B

3. B

4. A

5. B

Explanation:

Answer:

46.5%

Explanation:

mass of rock = 33g + 22g + 16g = 71g

% mass of zinc = (33/71) * 100%

= 0.465 * 100%

= 46.5%

stormwater pond .............

Coral reefs are believed to be the most diverse of any ecosystem. Hence it is most resilient to change due to its high diversity. Thus, option c is correct.

Coral reefs are sometimes referred to as the "rainforests of the sea" due to the richness of life that can be found there. The health of coral reefs is essential for around 25% of the ocean's fish. In the many crevices created by corals, fish and other species find shelter, food, breed, and raise their young.

More than 7,000 different kinds of fish, invertebrates, plants, sea turtles, birds, and marine mammals may be found here. Coral reefs offer chances for recreation, serve as a barrier against erosion and storm damage, and support local economies. They are also a source of fresh medications and food.

More than 500 million people rely on reefs for safety, income, and food. On and around reefs, fishing, diving, and snorkeling generate hundreds of millions of dollars for nearby businesses. Hence, option c is correct.

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Answer:

Part A: Protonation, Protonation

Part B: An elimination, A substitution

Explanation:

The dehydration ethanol and the condensation of ethanol all begin from the protonation of ethanol. This is because, water is a good leaving group. Thus, the protonation of the -OH moiety of the molecule is the first step in each mechanism.

The dehydration of ethanol is an elimination reaction because it involves the loss of H-O-H from the ethanol molecule while the condensation of ethanol is an substitution reaction because the -OH moiety is replaced by -OCH2CH3.

Answer:

plants can be used as fuel because plants can be burned so they compose the plants to make a liquid

Answer:

Well no because if metals lose electrons, any non-metal sources/items gain electrons from the metal.

Answer:

Metals tend to lose electrons and non-metals tend to gain electrons, so in reactions involving these two groups, there is electron transfer from the metal to the non-metal

Explanation:

Answer:

Explanation:

The main task here is that there are some missing gaps in the above question that needs to be filled with the appropriate answers. So, we are just going to do rewrite the answer below as we indicate the missing gaps by underlining them and making them in bold format.

SO; In the quantum-mechanical model of the hydrogen atom.

As the n level increases. the energy increases and thus levels are closer to each other. Therefore, the transition 3p→2s would have a greater energy difference than the transition from 4p→3p.

[tex]Energy \ and \ wavelength \ are \ inversely \ proportional , \ so \ the \ \mathbf{ 4p\to 3p} \ transition \ would[/tex][tex]produce \ a \ longer \ wavelength.[/tex]

Answer:

the answer is A

Explanation:

There are three kinds of plate tectonic boundaries: divergent, convergent, and transform plate boundaries.

This image shows the three main types of plate boundaries: divergent, convergent, and transform.

This image shows the three main types of plate boundaries: divergent, convergent, and transform. Image courtesy of the U.S. Geological Survey. Download image (jpg, 76 KB).

The Earth’s lithosphere, which includes the crust and upper mantle, is made up of a series of pieces, or tectonic plates, that move slowly over time.

A divergent boundary occurs when two tectonic plates move away from each other. Along these boundaries, earthquakes are common and magma (molten rock) rises from the Earth’s mantle to the surface, solidifying to create new oceanic crust. The Mid-Atlantic Ridge and Pacific Ring of Fire are two examples of divergent plate boundaries.

When two plates come together, it is known as a convergent boundary. The impact of the colliding plates can cause the edges of one or both plates to buckle up into a mountain ranges or one of the plates may bend down into a deep seafloor trench. A chain of volcanoes often forms parallel to convergent plate boundaries and powerful earthquakes are common along these boundaries.

At convergent plate boundaries, oceanic crust is often forced down into the mantle where it begins to melt. Magma rises into and through the other plate, solidifying into granite, the rock that makes up the continents. Thus, at convergent boundaries, continental crust is created and oceanic crust is destroyed.

Two plates sliding past each other forms a transform plate boundary. One of the most famous transform plate boundaries occurs at the San Andreas fault zone, which extends underwater. Natural or human-made structures that cross a transform boundary are offset—split into pieces and carried in opposite directions. Rocks that line the boundary are pulverized as the plates grind along, creating a linear fault valley or undersea canyon. Earthquakes are common along these faults. In contrast to convergent and divergent boundaries, crust is cracked and broken at transform margins, but is not created or destroyed.

Answer:

Explanation:

D or b

Sorry if wrong

Answer:

temporal

Explanation:

Answer:

Remain unchanged.

Explanation:

The total number of moles of liquid remain unchanged as the some moles of species B are added to the system because specie B that is added in the liquid phase is again restored after addition. If the specie B did not restored after addition to the liquid phase so the total number of moles increases in the liquid phase so that's why we can say that the liquid phase remain unchanged.

Answer:

oh ok

Explanation:

Answer:

Na(s) + Cl(g) = NaCl(s)

Explanation:

s means solid

g means gas

Metals are solid at room temperature.

Crystals are solids

Answer:

3.5hrs

Explanation:

Using the formula Speed = Distance/Time

Given

Speed = 8mi/hr

Distance = 28miles

Get the time

Time = Distance/Speed

Time = 28/8

Tme = 3.5hrs

Hence it will take 3.5hrs to travel

Answer:

A is brain, ND the fact that u don't know this stuff is a lil bit sad, u can so look it

Explanation:

Answer:

NH3

Explanation:

The "tri" attached to the trihydride means 3. So your hydrogen is going to have 3 atoms.

Answer:

All Answers only for the elite; Nomenclature of Covalent Compounds Quiz

You're Welcome :)

Explanation:

A-hydrochloric acid

B-sulfuric acid

C-dinitrogen pentoxide

C- IBr3

A- carbon dioxide

B- Start the name with hydro-.

B- H3BO3

C-sulfur trioxide

A- NH3

B- The chemical name ends with “hydroxide.”

Answer:

nonmtals:

Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process. It is also used for manufacturing of steel and provides high temperature in metal fabrication process. ...

Nonmetals used in fertilizers: Fertilizers contain nitrogen. It helps in plant growth. ...

Nonmetals used in crackers

Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Seven non-metals exist under standard conditions as diatomic molecules: H2(g)

Explanation:

Answer:

a nonpolar covalent

b polar

c nonpolar

d ionic

e. polar

f. ionic

g polar

h polar

Explanation:

Answer:

The correct answer is B. equals mass divided by volume

Answer:

They were right, I got all the questions right. If you need answers I got them I guess?

Explanation:

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point F)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

Our initial mmoles of OH⁻ would not change through all the titration.

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

The pH value is given by 14 less the logarithm of the hydronium ion

concentration or the hydrogen ion concentration in the solution.

Responses (approximate values):

The pH values are;

A) 13.11

B) 12.91

C) 12.71

D) 12.42

E) 11.54

F) 7

G) 2.469

H) 1.884

A) Concentration of the KOH = 0.129 M

Amount of HCl added = 0.00 ml

The pH = -log[H⁺] = 14 - pOH

pOH = -log[OH⁻]

Which gives;

pH = 14 - (-log[OH⁻] )

B) Volume of acid added = 7.00 mL = 0.007 L

Concentration of the acid = 0.258 M HCl

Number of moles of acid, H⁺ = 0.007 × 0.258 moles = 0.001806 moles

Number of moles of KOH remaining, OH⁻= 0.05 × 0.129 - 0.001806 = 0.004644

Number of moles of OH⁻ = 0.004644 moles

[tex]Concentration, \ [OH^-] = \dfrac{0.004644 \, moles}{0.05 7 \, L} \approx \mathbf{ 0.0815 \, M}[/tex]

C) 12.5 mL HCl contains, 0.0125 × 0.258 moles = 0.003225 moles

OH⁻ remaining = 0.05 × 0.129 - 0.003225 = 0.003225 moles

[tex]Concentration \ of \ [OH^-]= \dfrac{0.003225\, moles}{(0.05 + 0.0125) \, L} = \mathbf{0.0516 \, M}[/tex]

D) 18.0 mL HCl contains, 0.018 × 0.258 moles = 0.004644 moles

OH⁻ remaining = 0.05 × 0.129 - 0.004644 = 0.001806 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0.001806\, moles}{(0.05 + 0.018) \, L} \approx \mathbf{0.0266\, M}[/tex]

E) 24.0 mL HCl contains, 0.024 × 0.258 moles = 0.006192 moles

OH⁻ ion remaining = 0.05 × 0.129 - 0.006192 = 0.000258 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0.000258\, moles}{(0.05 + 0.024) \, L} \approx \mathbf{0.0035\, M}[/tex]

F) 25.0 mL HCl contains, 0.025 × 0.258 moles = 0.00645 moles

[OH⁻] remaining = 0.05 × 0.129 - 0.00645 = 0 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0\, moles}{(0.05 + 0.024) \, L} \approx 0\, M[/tex]

Therefore;

Number of moles of KOH = 0, or the solution is neutralized

[OH⁻] = [H⁺]

Which gives;

G) 26.0 mL HCl contains, 0.026 × 0.258 moles = 0.006708 moles

[OH⁻] remaining = 0.05 × 0.129 - 0.006708 = -0.000258 moles

Therefore

Number of moles of H⁺ = 0.000258

[tex]\mathbf{Concentration} \ of \ \mathbf{[H^+] }= \dfrac{0.000258\, moles}{(0.05 + 0.026) \, L} \approx 0.003395\, M[/tex]

H) 29.0 mL HCl contains, 0.029 × 0.258 moles = 0.007482 moles

H⁺ remaining = 0.007482 - 0.05 × 0.129 = 0.001032 moles

Therefore

Number of moles of H⁺ = 0.001032

[tex]Concentration \ of \ [H^+] = \mathbf{ \dfrac{0.001032\, moles}{(0.05 + 0.029) \, L}} \approx 0.01306\, M[/tex]

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Answer:

because all chemical reaction involve the rearrangement of atoms, the mass is an electron is negatively charged, it has a mass approximately 1/2000 that of a, Atoms of an element that contain the same number of protons but a different Copy and complete the following tables concerning the isotopes of silicon SI-28.

Explanation:

An alloy always has two or more elements/metals in it. An alloy always has at least metal in it.

Steels have iron and carbon in it. Bronze is copper and tin, or aluminum, or manganese, or zinc. German silver isn't silver at all and contains copper, zinc, and nickel.

Hope this helps!

Answer:

a. 0.68 M NaCl solution

Explanation:

The tonicity of a solution can either be hypotonic, hypertonic and isotonic. Hypertonic solution is a solution which possesses a higher concentration of solute in relation to another solution.

According to this question, a solution is said to contain 0.15 M NaCl solute. This means that a solution that will be hypertonic to this solution will have a much more concentration of solute, which based on the options provided is the 0.68 M NaCl solution.