Pleeeeasee someone who’s good at chemistry?! 10 grade

ASAP
I’ll give points, just help please

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

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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One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.

Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.

We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.

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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.

Explanation:

since 5g saturates 12.5g of water at 10°c

so......x would saturate 50g of water at 0°c

then you can cross multiply

Answer:

Tetrahedral molecules are normally spy hybridized.

Explanation:

Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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Answer:

Model C has more double the oxygen than A. I think it is B

Explanation:

Answer:

16.7%

Explanation:

40 ml of salt water + 200 ml of solution = 240 ml

40/240 = 4/24 = 1/6=16.7%

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: the answer is 0.074Hz

Explanation:

Given, (In air)

Velocity V=340m/s

Frequency f=20,000Hz

Wavelegth λ=?

V=f.λ

λ=

F

V

​

=

20,000

340

​

=0.017Hz

Also, Given (in Water)

Velocity, V=1480m/s

Frequencyf=20,000Hz

wavelength, λ=?

V=F.λ

λ=

F

V

​

=

20,000

1480

=​0.074Hz

Answer:

frequency

Explanation:

frequency is velocity/ wavelength

340/1.36

250

Answer:

0.00132moles

Explanation:

1000ml of NaOH contain 0.110 moles

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

= 0.00132 moles

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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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.

The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Cr

But:

1 mole of Cr = 52 g

Thus, we can say that:

6.02×10²³ atoms = 52 g of Cr

With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:

6.02×10²³ atoms = 52 g of Cr

Therefore,

5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³

5.15×10²¹ atoms = 0.44 g of Cr

Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

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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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Answer:

Explanation:

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

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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Explanation:

the investigation lasts for 7 days !

hope this helps you.

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].

In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.

When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

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Answer:

Molecules vibrate in a fixed position.

Explanation:

first one is just wrong.

third is liquid

fourth is gas

Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]

Thus, the density of the calcite sample will be:

[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

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Regards!