The empirical formula of a sample of gold bromide was found by adding aluminum metal to an aqueous solution of gold bromide, displacing gold metal.

a. True
b. False

Answer:

Numbers after atoms are subscripts. Numbers before atoms are coefficients.

Equation: 2Na +Cl2 -> 2NaCl

Given: 6.8 grams of Na  

Unknown: moles of NaCl

Set up: [tex]6.8g Na *\frac{1 mole Na}{22.99 g Na}*\frac{2 moles NaCl}{2 moles Na} =.2957 moles NaCl[/tex]

Explanation:

We know that the reactants are Na and Cl and that what is being produced, the products, is NaCl.

Na + Cl -> NaCl

CAREFUL! Cl is a diatomic ion, is it gets a subscript of 2! This means there need to be 2 cl atoms on the right side as well. But now there are 2 Na on the right side. So we multiply Na by 2 on the left so that both sides have two Na atoms.

2Na +Cl2 -> 2NaCl

B)

We know that there are 6.8 grams of Na. This is our given.

What do we want to find out? We want to find how many moles of NaCl there are, so that is our unknown.

C) A mole ratio is exactly what it sounds like. From our balanced equation, we can see that there are 2 moles of Na  and 2 moles of NaCl. In other words, there are 2 moles of Na for every 2 moles Nacl. In notation, this is [tex]\frac{2 mole Na}{2 Mole NaCl}[/tex] or 2:2. You can simplify to 1:1 ratio, but as equations get more complex I find it easier to not. It depends on what your teacher wants and what you find easier.  

D)

By set up I assume they mean stoichiometry (chemistry conversion) . First I'm going to split up the set up so that it is easier to understand.

We start with the given and convert that into moles Na. We can do this by remembering that there are 22.99 grams for every 1 mole Na:

6.8 g Na* [tex]\frac{1 mole Na}{22.99g Na} = .2957 moles Na[/tex]

Now we can use that mole ration we found in part C to convert moles of Na into moles of NaCl:

[tex].2957 moles Na *\frac{2 moles NaCl}{2 moles Na}= .2957[/tex]

Usually, we will put these together to get

[tex]6.8g Na *\frac{1 mole Na}{22.99 g Na}*\frac{2 moles NaCl}{2 moles Na} =.2957 moles NaCl[/tex]

Tip: in Stoichiometry, it is important to remember that your units should cancel out until you only have the units you want. If there are moles of Na in the numerator, there should be moles of Na in the denominator of the following fraction. If there are grams Na in the numerator, there should be grams Na in the denominator and so and so on until you are only left with the unit you want, which, in this case, is moles of NaCl.  

Answer:

9th:08473 kJ/mol

Explanation:

HAHAHAHAHAHAHAHAHAHA

Answer:

:N:::N:

Explanation: 10 valence electrons betwwen the 2

need an octet

and N needs 5 for a neutral formal charge

An acceptable Lewis structure for a diatomic nitrogen molecule is

:N:::N: .Option C is correct. This is further explained below.

Lewis structure for a diatomic nitrogen molecule simply refers to the triple bond exists between two nitrogen atoms in the N2 Lewis structure.

In conclusion, The nitrogen molecule is diatomic, which means it is made up of two atoms of the same element.

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Answer: 0.25

Explanation: M = moles of solute / volume of solution (L)

First, you need to calculate the number of moles using the given mass and the molar mass.

15 g74.692 g/mol≈0.20 mol

Then, you can solve for molarity using the molarity equation. The answer should be reported to two significant figures.

M=0.20 mol/ 0.800 L = 0.25 M

The molar concentration of  NiO solution is 0.251 M.

Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.

The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.

In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.Considering the given example, molar concentration of NiO   is found out as follows,

molar concentration=15/74.69 ×1/0.8= 0.251 M.

where 74.69 is molar mass of NiO.

Thus, the molar concentration of NiO is 0.251 M.

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Answer:

The answer is A.Newton's First law

Answer: 15.0 g of [tex]CCl_4[/tex] will be formed upon the complete reaction of 27.7 grams of chlorine gas with excess carbon disulfide

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Cl_2=\frac{27.7g}{71g/mol}=0.390moles[/tex]

The balanced chemical reaction is:

[tex]CS_2(g)+4Cl_2(g)\rightarrow CCl_4(l)+2SCl_2(s)[/tex]

[tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]CS_2[/tex] is the excess reagent.

According to stoichiometry :

4 moles of [tex]Cl_2[/tex] give=  1 mole of [tex]CCl_4[/tex]

Thus 0.390 moles of  will give = [tex]\frac{1}{4}\times 0.390=0.0975moles[/tex] of [tex]CCl_4[/tex]

Mass of [tex]CCl_4=moles\times {\text {Molar mass}}=0.0975\times 154=15.0g[/tex]

Thus 15.0 g of [tex]CCl_4[/tex] will be formed upon the complete reaction of 27.7 grams of chlorine gas with excess carbon disulfide

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:

8 neutrons

A is the number of neutrons plus protons in the nucleus. However, we already know that there are 7 protons. Therefore, there must be 8 neutrons in the nucleus to add up to 15. Table 3.

Element Symbol Number of Neutrons

carbon 146C 8

nitrogen 147N 7

157N 8

oxygen 168O 8

Explanation:

Answer:

3

Explanation:

1 Calcium atom and two chlorine atoms

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:

exothermic means that energy was released during the chemical reaction

Explanation:

Answer:

b) because the positive charge is on primary and a tertiary carbon

Explanation:

The allylic carbon atom carries positive charge when primary carbon has positive charge. It forms an ion with tertiary carbon atom to form resonance. In secondary carbocation the positive charge is on secondary carbon atom. The atom are attracted and electrons form carbon -  carbon sigma.

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:

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:

the answer is B

Explanation:

Answer:

answer is number 2.

Explanation:

hope this helps

Answer:

Explanation:

That's correct. Once Aluminum becomes an ion, it is very hard to force it to take back its electrons. Only a few elements can do it. Iron is not one of them.

Answer:

Noble gases

Explanation:

Neon and argon have 8 electrons in their outermost shells. The elements which have 8 electrons in the outermost shell or in the valence shell,are known as the noble gases.

It's the third one or Noble gases.

Answer:

oh ok

Explanation:

Answer:

Nice

Explanation:

Answer:

The answer is A.

Explanation:

Answer:

Explanation:

I believe it is a

Answer: The mass of [tex]CO_2[/tex] and [tex]H_2O[/tex] produced are 336.6 g and 183.6 g respectively.

Explanation:

The combustion reaction between propane and oxygen leads to formation of carbon dioxide and water.

Law of Conservation of mass states that the mass will remain constant for a balanced equation. This is carried out when the total number of atoms on reactant side is same as the total number of atoms on the product side. Thus the equation must be balanced.

[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]

a) 1 mol of propane produces = 3 moles of [tex]CO_2[/tex]

Thus 2.55 mol of propane produces = [tex]\frac{3}{1}\times 2.55=7.65 moles of [tex]CO_2[/tex]

mass of [tex]CO_2=moles\times {\text {molar mass}}=7.65mol\times 44g/mol=336.6g[/tex]

b) 1 mol of propane produces = 4 moles of [tex]H_2O[/tex]

Thus 2.55 mol of propane produces = [tex]\frac{4}{1}\times 2.55=10.2 moles of [tex]H_2O[/tex]

mass of [tex]H_2O=moles\times {\text {molar mass}}=10.2mol\times 18g/mol=183.6g[/tex]

The mass of [tex]CO_2[/tex] and [tex]H_2O[/tex] produced are 336.6 g and 183.6 g respectively.

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:

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:

2.26 × 10²⁴ atoms Zn

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

3.75 mol Zn (Zinc)

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2.25825 × 10²⁴ atoms Zn ≈ 2.26 × 10²⁴ atoms Zn

Answer: D

Explanation: Today, we know that the continents rest on massive slabs of rock called tectonic plates. They are constantly in movement and they are responsible for several natural disasters. Approximetly 335 million years ago, during the late Paleozoic and early Mesozoic eras, the world was one big mass, called Pangea, which eventually would start to break apart about 175 million years ago. At the center of the Earth is the core, which has two parts. The solid, inner core of iron has a radius of about 760 miles (about 1,220 km), according to NASA. It is surrounded by a liquid, outer core composed of a nickel-iron alloy.

I hope it helps

Answer:

the answer is D

Explanation:

Answer:

8.8mol/l

Explanation:

but I'm not percent sure.

Answer:

I don't think so for now, there hasn't been much research done with the vaccine and children so we really don't know how children's bodies will react to it. Once more studies are done with the vaccine and children and is proven safe then yes, kids should use it in order to attend public school.

Answer:

The atomic number increases across a period and down the group or family. The atomic mass generally increases with atomic number with a few exceptions.

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)  The maximum mass of sulfur trioxide that can be formed is 31.4 grams

b) The FORMULA for the limiting reagent is [tex]O_2[/tex]

c) Mass of excess reagent remains is 4.8 grams

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} SO_2=\frac{29.9g}{64g/mol}=0.467moles[/tex]

[tex]\text{Moles of} O_2=\frac{6.26g}{32g/mol}=0.196moles[/tex]

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]  

According to stoichiometry :

1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]

Thus 0.196 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.196=0.392moles[/tex]  of [tex]SO_2[/tex]

Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]SO_2[/tex] is the excess reagent as (0.467-0.392) = 0.075 moles or [tex]0.075mol\times 64g/mol=4.8g[/tex] are left.  

As 1 mole of [tex]O_2[/tex] give = 2 moles of [tex]SO_3[/tex]

Thus 0.196 moles of [tex]O_2[/tex] give =[tex]\frac{2}{1}\times 0.196=0.392moles[/tex]  of [tex]SO_3[/tex]

Mass of [tex]SO_3=moles\times {\text {Molar mass}}=0.392moles\times 80g/mol=31.4g[/tex]

Thus 31.4 g of [tex]SO_3[/tex] will be produced from the given masses of both reactants.