Write the balanced half-reactions and an overall cell reaction below. Be sure to include states of matter.

Answer:

[tex]\boxed {\boxed {\sf 0.757 \ atm}}[/tex]

Explanation:

We are asked to find the pressure of a gas given a change in volume. Since the temperature remains constant, we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume is inversely proportional to the pressure. The formula for this law is:

[tex]P_1V_1= P_2V_2[/tex]

Initially, the oxygen gas occupies a volume of 18.7 liters at a pressure of 1.19 atmospheres.

[tex]1.19 \ atm * 18.7 \ L = P_2V_2[/tex]

The gas expands to a volume of 29.4 liters, but the pressure is unknown.

[tex]1.19 \ atm * 18.7 \ L = P_2 * 29.4 \ L[/tex]

We are solving for the new pressure, so we must isolate the variable [tex]P_2[/tex]. It is being multiplied by 29.4 liters. The inverse operation of multiplication is division. Divide both sides of the equation by 29.4 L.

[tex]\frac {1.19 \ atm * 18.7 \ L}{29.4 \ L} =\frac{ P_2 * 29.4 \ L}{29.4 \ L}[/tex]

[tex]\frac {1.19 \ atm * 18.7 \ L}{29.4 \ L} =P_2[/tex]

The units of liters cancel.

[tex]\frac {1.19 \ atm * 18.7 }{29.4 } =P_2[/tex]

[tex]\frac {22.253}{29.4 } \ atm = P_2[/tex]

[tex]0.7569047619 \ atm =P_2[/tex]

The original measurements all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 9 in the ten-thousandth place to the right of this place tells us to round the 6 up to a 7.

[tex]0.757 \ atm \approx P_2[/tex]

The pressure of the gas sample is approximately 0.757 atmospheres.

According to Boyle's law, for a given mass of ideal gas, pressure of gas is inversely proportional to the volume of gas, Provided the Temprature remains constant.

[tex]\implies \sf P_1 V_1 = P_2 V_2 \\ [/tex]

[tex]\implies \sf 1.19 \times 18. 7= P_2 \times 29.4 \\[/tex]

[tex]\implies \sf 22.253= P_2 \times 29.4 \\[/tex]

[tex]\implies \sf P_2 = \dfrac{22.253}{29.4} \\[/tex]

[tex]\implies \underline{ \red{\boxed{ \bf P_2 \approx0.756 \: atm }}} \\[/tex]

Answer:

[Kr] 5s²

Explanation:

From the question given above, the following data were obtained:

Atomic number of strontium (Sr) = 38

Electronic configuration =?

Next, we shall determine the electronic configuration of the noble gas element before strontium (Sr).

The noble gas element before strontium (Sr) is krypton (Kr). Thus, the electronic configuration of krypton (Kr) is given below:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Finally, we shall determine the electronic configuration of strontium (Sr). This can be obtained as follow:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Sr (38) =>?

Sr (38) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶ 5s²

But

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Therefore,

Sr (38) => [Kr] 5s²

Answer:

» Water evaporating to gas.

» Solid subliming to gas.

» Water freezing to solid.

» Solid melting to liquid.

Explanation:

[tex]{ \underline{ \blue{ \sf{christ \:† \: alone }}}}[/tex]

We want to find the value of t such that the velocity vector makes an angle of 45° with both axes.

We found that:

t = (2/3)*(b/c)

The velocity vector makes an angle of 45° with the x and y axes.

We know that the position vector is:

r = r=b*t²i + c*t³j

Remember that the versor "i" corresponds to the x-component, and the versor "j" corresponds to the y-component, then:

r = r=b*t²i + c*t³j  = (b*t², c*t³)

The velocity vector is the vector that we get when we differentiate the position one, remember that if:

f(x) = a*x^n

then

f'(x) = n*a*x^(n - 1)

Using this, we can find that the velocity vector is:

v = (2*b*t,  3*c*t²)

Now we want to know, when does the velocity vector make an angle of 45° with the x and y axes.

Let's think of the vector as the hypotenuse of a triangle rectangle, where the x-component is the adjacent cathetus, and the y-component is the opposite cathetus. (so the angle is measured counterclockwise from the x-axis)

We have the trigonometric equation:

tan(a) = (opposite cathetus)/(adjacent cathetus)

So now we can replace these things with the known ones:

a = 45°

opposite cathetus = y-component = 3*c*t²

adjacent cathetus = x-component = 2*b*t

So we will get:

tan(45°)  =  (3*c*t²)/( 2*b*t)

1 = (3/2)*(c/b)*t

Now we can solve this for the variable, t.

1*(2/3)*(b/c) = t

t = (2/3)*(b/c)

We can conclude that at the time:

t = (2/3)*(b/c)

The velocity vector makes an angle of 45° with the x and y axes.

You can read more about vectors in:

https://brainly.com/question/10841907

Answer:

B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.

Explanation:

There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.

One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.

Answer:

See explanation and image attached

Explanation:

The reaction of HBr with 1-butene is an addition reaction. The HBr adds across the double bond to yield a saturated halogenoalkane.

In the absence of peroxides, the reaction proceeds in accordance with Markovnikov rule which states that;''the negative part of the addendum, is attached to the carbon atom with the least number of hydrogen atoms attached.'' This occurs in the first reaction shown in the image.

In the presence of peroxides, the reaction proceeds in an anti-Markovnikov manner to yield the product shown in the second reaction.

Answer:

pH = 2

Explanation:

Trioxonitrate (v) acid​ is also known as nitric acid (HNO₃) and is one of the strong acid set which when dissolved in water, ionizes 100%. That is,

0.01M HNO₃ => 0.01M H⁺ + 0.01M NO₃⁻ => pH = -log[H⁺] = -log(0.01) = -(-2) = 2

Answer:

1:Part A.

[tex]\bold{42.2 g C_{12}H_{22}O_{11} \:in \:528 g H₂O}[/tex]

Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]

=[tex]\frac{42.2}{528}*100\%=\bold{\underline{7.99\: or \:8\%}}[/tex]

Part B.

[tex]\bold{198\:m g\: C_{6}H_{12}O_{6} \:in\:4.71 g\: H₂O}[/tex]

mass of solute: 198mg

mass of solvent :4.71g=4710g

Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]

=[tex]\frac{198}{4710}*100\%=\bold{\underline{4.20\%}} [/tex]

Part C.

[tex]\bold{8.85 g NaCl \:in \:190 g\: H₂O}[/tex]

Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]

=[tex]\frac{8.85}{190}*100\%=\bold{\underline{4.66\%}}[/tex]

Answer:

It will help you !!!!!!!!!!

Answer:

D. beta particle

Explanation:

Number of protons increases from 38 to 39 indicating beta decay (only one proton up from parent isotope to daughter isotope) Also atomic mass (on top of an isotope), 90 stays the same as beta particle is very small.

Answer:

True

Explanation:

Aromatic compounds undergo substitution rather than addition reactions because the aromatic structure is maintained.

Electrophilic aromatic substitution begins with attack of the electrophile on the aromatic ring to yield a delocalized intermediate called the arenium intermediate. Loss of hydrogen from this intermediate yields the final product.

Answer:

i believe its 42

Explanation:

Answer:

19.2 g

Explanation:

Step 1: Write the balanced equation

P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄

Step 2: Calculate the moles corresponding to 5.29 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

5.29 g × 1 mol/18.02 g = 0.294 mol

Step 3: Calculate the theoretical yield of phosphoric acid, in moles

The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol

Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

0.196 mol × 97.99 g/mol = 19.2 g

Explanation:

Because molarity is mol/L, we'll have to convert 17g to mol.

After obtaining the mol, we'll divide that by the volume to obtain Molarity.

[tex]\\ \sf\longmapsto NH_3[/tex]

[tex]\\ \sf\longmapsto 14u+3(1u)[/tex]

[tex]\\ \sf\longmapsto 14u+3u[/tex]

[tex]\\ \sf\longmapsto 17u[/tex]

[tex]\\ \sf\longmapsto 17g/mol[/tex]

Moles of Ammonia:-

[tex]\boxed{\sf No\:of\:moles =\dfrac{Given\:mass}{Molar\:mass}}[/tex]

[tex]\\ \sf\longmapsto No\:of\:moles=\dfrac{17}{17}[/tex]

[tex]\\ \sf\longmapsto No\;of\:moles=1mol[/tex]

We know

[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\: Solution\:in\:L}}[/tex]

[tex]\\ \sf\longmapsto Molarity=\dfrac{1}{0.50}[/tex]

[tex]\\ \sf\longmapsto Molarity=2M[/tex]

Answer:

2Ag(s) + 2H²S(g) + O2(g) ➡️ Ag2S(s) + 2H2O(g)

Explanation:

Sorry for my typo. but you understand

Answer:

0

Explanation:

answer from gauth math

В

an echo is caused by the fact that waves can be reflected by solid surfaces, this is due to the dynamic pattern of rarefactions and air seals near the reflecting surface

Answer:

pH = 10.20

Explanation:

The HClO reacts with NaOH as follows:

HClO + NaOH → H2O + NaClO

Where HClO and NaOH react in a 1:1 reaction.

As the concentration of both reactions is the same and the reaction is 1:1, to reach equivalence point are required the same 25.0mL.

And the NaClO produced decreases its concentration in 2 because the volume is doubled.

The concentration of NaClO is: 0.150M / 2 = 0.075M

The equilibrium of NaClO is:

NaClO(aq) + H2O(l) ⇄ HClO(aq) + OH-(aq)

Where Kb of reaction is 1.0x10⁻¹⁴ / Ka =

1.0x10⁻¹⁴ / 3.0x10⁻⁸ = 3.33x10⁻⁷ = [HClO] [OH-] / [NaClO]

[NaClO] = 0.075M

As both HClO and OH- comes from the same equilibrium,

[HClO] = [OH-] = X

Where X is the reactoin coordinate

Replacing:

3.33x10⁻⁷ = [X] [X] / [0.075M]

2.5x10⁻⁸ = X²

X = 1.58x10⁻⁴M = [OH-]

pOH = -log [OH-]

pOH = 3.80

pH = 14 - pOH

Answer:

I think b acid makers

Explanation:

hope it will help you

Answer:

Explanation:

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

So ,

[tex]\\ \Large\sf\longmapsto 250\times 15\%[/tex]

[tex]\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}[/tex]

[tex]\\ \Large\sf\longmapsto 37.5g[/tex]

37.5g of NaCl present in 250g of solution.

Answer:

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

➡250 × 15%

➡250×15/100

➡37.5g

37.5g of NaCl present in 250g of solution.

Answer:

C2H5P

is the right answer

Answer:

C: The clothing company used technological design to create a product that helped aid researchers conducting a scientific investigation.

Explanation:

I just did the assignment on EDGE2020 and it's 200% correct!  

Also, heart and rate if you found this answer helpful!! :) (P.S It makes me feel good to know I helped someone today!!)  :)  

Answer:

The Avogadro's number

Explanation:

the number is 6.02214x10 to the 23 power

Answer:

Hg(NO3)2

Explanation:

Hg occurs both as Hg(I) and Hg(II) but nitrate is -1, not -2 We NORMALLY use the smallest set of subscripts but in the case of Hg(I) compounds, most of them are dimeric. Hg2Cl2, Hg2(NO3)2, etc. Note that there also is a mercury(II) nitrate; i.e., Hg(NO3)2. Therefore, although the empirical formula for mercury(I) nitrate is HgNO3, the molecular formula is Hg2(NO3)2 anf I was always encouraged to write the molecular formula instead of the empirical formula.

Hope this helps! Feel free to shoot me any questions <3

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___Cu+² __|_2OH-____|

|Initial concentration(M)|___0__|_0______|

|Change in concentration(M)|_+S |__+2S__|

|Equilibrium concentration(M)|_S _|2S___|

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

[tex]s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7} [/tex]

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

[tex]Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5} [/tex]

I hope I helped you^_^