7. What's the structure of methane, CH,? Is it polar or non-polar?
O A. Trigonal planar, polar
O B. Tetrahedral, polar
O C. Trigonal planar, non-polar
O D. Tetrahedral, non-polar

Answer:

Fe is the excess reactant

Reaction yields are the amount of the reactant and the products of a chemical reaction. In the reaction between iron and oxygen, iron is the excess agent.

The reactant present in an extra quantity than the other reactant in a chemical reaction which reacts with the limiting reactant is called an excess reactant.

In a reaction mixture, the excess reactant is present even when the limiting agent is completely consumed.

The chemical reaction between iron and oxygen is shown as,

[tex]\rm 4 Fe(s) + 3 O_{2}(g) \rightarrow 2 Fe_{2}O_{3}[/tex]

From the reaction, it can be said that oxygen is a limiting reagent that limits the formation of iron (iii) oxide.

Therefore, iron is an excess reactant.

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Answer:

the small intestine where nutrients are absorbed.

3HCl + Al(OH)3 > AlCl3 + 3H20

mol = conc × v

= 0.367 × 0.216

= 0.0792 mol HCl

3 mol HCl = 1 mol Al(OH)3

0.0792 mol HCl = x

x = 0.0792/3 × 1

= 0.0264 mol Al(OH)3

Al(OH)3 = 27 + 3(16 +1) = 78 g/mol

mass = mol x molar mass

= 0.0264 × 78

= 2.0592 g

I don't know if it's correct

Answer:

Molecules vibrate in a fixed position.

Explanation:

first one is just wrong.

third is liquid

fourth is gas

Answer:

310K

Explanation:

Rearrange PV=nRT to get T=PV/nR

T=(2.0L)(0.75atm)/(0.5mol)(0.08206)

=36.5 or 37

add 273 for K to get

310

Answer:

230293

Explanation:

4fe(lll) + 3o2 2o2 2Fe 2O3 + 230293

Answer:

Explanation:

You have to find the mass in the mole before the molarity and then in the mole.

Explanation:

gravity

kinetic energy

less than

friction resists the ball's movement

2.0 m/s

change its direction

speed

newton's first law

dump truck

position

Answer:

Q1 - Gravity

Q2 - Kinetic Energy

Q3 - Less than

Q4 - Friction resists the ball's forward motion

Q5 - 2 m/a

Q6 - Change its speed

Q7 - Speed

Q8 - Second law

Q9 - a dump truck

Q10 - motion

I am not sure about some answers but I answered all questions.

Work is said to be done if the force applied to a body causes the body to move through a distance.

Student B used more work and power

Workdone = Force * Distance

For student A:

Force = 3N

distance = 5meters

Time taken = 10 secs

Workdone by Student A = 3 * 5

Workdone by student A = 15Nm

Power used up = workdone/time

Power used up = 15/10

Power used up = 1.5 Watts

For student B:

Force = 6N

distance = 5meters

Time taken = 5 secs

Workdone by Student A = 6 * 5

Workdone by student A = 30Nm

Power used up = workdone/time

Power used up = 30/5

Power used up = 6 Watts

This shows that student B used more work and power

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The answer would be 4.

The agencies are:

Department of Agriculture's Forest Service (AFS)

Department of the Interior's Bureau of Land Management (BLM)

Fish and Wildlife Service (FWS)

National Park Service (NPS)

These four agencies cover around 95% of the land and are the only federal agencies.

I learned about these agencies in school.

Answer:

Approximately [tex]83[/tex] minutes.

Explanation:

Look up the relative atomic mass of [tex]\rm Ag[/tex]: [tex]M({\rm Ag}) = 107.868\; \rm g \cdot mol^{-1}[/tex].[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 1.6987 \times 10^{4}\; \rm C \end{aligned}[/tex]/.

Avogadro's number: [tex]N_A \approx 6.02 \times 10^{23}\; \rm mol^{-1}[/tex].

Elementary charge: [tex]e \approx 1.602 \times 10^{-19}\; \rm C[/tex].

Calculate the quantity of [tex]\rm Ag[/tex] atoms to reduce:

[tex]\begin{aligned}& n({\rm Ag}) \\ &= \frac{m({\rm Ag})}{M({\rm Ag})} \\ &= \frac{19\; \rm g}{107.868\; \rm g \cdot mol^{-1}} \\ & \approx 0.176\; \rm mol\end{aligned}[/tex].

By the equation, it takes one electron to reduce every [tex]\rm Ag[/tex] atom. Thus, the number of electrons required to reduce [tex]0.176\; \rm mol[/tex] of [tex]\rm Ag\![/tex] atoms would be:

[tex]n(e) = n({\rm Ag}) \approx 0.176\; \rm mol[/tex].

[tex]\begin{aligned}N(e) &= n(e) \cdot N_{A}. \\ &\approx 0.176\; \rm mol \times 6.02 \times 10^{23}\; \rm mol^{-1} \\ & \approx 1.06 \times 10^{23}\end{aligned}[/tex].

Calculate the amount of charge (in coulombs) in that many electrons:

[tex]\begin{aligned}Q &= e\, (n(e)) \\ &\approx 1.602 \times 10^{-19}\; \rm C \times 1.06 \times 10^{23} \\ &\approx 16987.1 \; \rm C \end{aligned}[/tex].

A current of [tex]1\; \rm A[/tex] carries a charge of [tex]1\; \rm C[/tex] every second. Thus, the amount of time required for this current to carry that much electron would be:"

[tex]\begin{aligned}t &= \frac{Q}{I} \\ &\approx \frac{16987.1\; \rm C}{3.4\; \rm A} \\ &\approx 83.3\; \rm s \\ &\approx 5.00\times 10^{3}\; \rm s \\ &\approx 83\; \text{minutes} \end{aligned}[/tex].

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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From the information available in the question, the concentration of the acid is 0.21 M.

H3PO4(aq) + 3NaOH(aq) -----> Na3PO4(aq) + 3H2O(l)

Volume of acid(VA) = 51.0 mL

Concentration of acid (CA) = ?

Volume of base (VB) = 26.2 mL

Concentration of base (CB) = 1.25 M

Number of moles of acid (NA) = 1

Number of moles of base (NB) = 3

Using the formula;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA =  CBVBNA/VANB

CA =  1.25 M × 26.2 mL × 1/51.0 mL × 3

CA = 0.21 M

The concentration of the acid is 0.21 M.

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Answer:

Yes it is B,D.

Explanation:

Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups.

Explanation:

Mole ratio of NaOH : HCl = 1: 1

Moles of NaOH used = 0.1 mol/1000 cm3 × 27.20 cm3

= 2.72 × 10^-3 mol

Therefore moles of HCl used is also 2.72 × 10^-3 mol

So concentration of HCl can be found by dividing the no.of moles of HCl by the volume of HCl as follows

2.72 × 10^-3 mol/ 5cm3

1000cm3 = 1dm3

Therefore,

1cm3 = 1/1000 dm3

5cm3 = 5/1000 dm3

HCl conc. = 2.72 × 10^-3 mol/ 5×10^-3 dm3

= 0.544 moldm-3

Answer: The following information should be included;

What is the difference between the greenhouse effect, climate change and global warming?

What proof do we have that climate change is happening?

Why is it happening?

Answer:

0.00132moles

Explanation:

1000ml of NaOH contain 0.110 moles

12ml of NaOH contain (12*0.110)/1000 moles

= 0.00132 moles

Answer:

Ptotal=P1+P2+… +Pn. + P nExplanation:

its c

The pH of the standard hydrogen electrode that has electrode potential of -0.128 V  is 4.3.

The equation of the hydrogen electrode is;

2H^+(aq) + 2e ⇄ H2(g)

The standard electrode potential of hydrogen is 0.00 V

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Now;

E°cell = 0.00 V

n = 2

Q = 1/[H^+]

-0.128 = 0.00 - 0.0592/2 log  1/[H^+]

-0.128 = 0.00 - 0.0296 log 1/[H^+]

 -0.128 =  - 0.0296 log 1/[H^+]

-0.128/  - 0.0296  =  log 1/[H^+]

1/[H^+] = Antilog (4.32)

[H^+] = 4.79 × 10^-5

Now;

pH = -log[H^+]

pH = -log (4.79 × 10^-5)

pH = 4.3

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Answer:

*The coastal areas experience less variation in temperature as compared to land. ... It means that water takes more time to get hot or cold then land. Therefore areas near sea or ocean have more moderate temperature in the day time and less cooler at night

Explanation:

When radioactive decay occurs, the original nucleus splits into daughter nuclei and the resulting nucleus is more stable than the original nucleus. The nucleus can be of a different element than the original.

Unstable nuclei often undergo radioactive decay. In a radioactive decay, the unstable nucleus is broken up into other nuclei. Usually, the nuclei formed during radioactive decay are smaller in mass compared to the original nucleus.

Also, the resulting nucleus is more stable than the original nucleus. The nucleus can be of a different element than the original.

The effect is that the nucleus changes into the nucleus of one or more other elements. These daughter nuclei have a lower mass and are more stable (lower in energy) than the parent nucleus.

This  problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.

Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho =\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:

[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]

And finally the required density in g/cm³:

[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]

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Explanation:

the investigation lasts for 7 days !

hope this helps you.