An inflated balloon is left
outside overnight.
Initially, it has a volume
of 1.74 L when the
temperature is 20.2°C and
the pressure is 1.02 atm.
At what temperature will
the balloon have a volume
of 1.56 L if the pressure
falls to 0.980 atm?
*Show all of your work*

Answer:

0.60 moles of atoms of carbon

Explanation:

Step 1: Given data

Step 2: Calculate the number of carbon atoms in the given sample

According to the chemical formula of the compound, the molar ratio of C to O is 2:6, that is, there are 2 moles of atoms of C every 6 moles of atoms of O. The number of moles of atoms of C is:

1.8 mol O × 2 mol C / 6 mol O = 0.60 mol C

Answer:

ecosystem i think

Answer: are there any answer choices? If not I think MOTION

Explanation: hope that’s right, have a great day!! :)

Answer:

copper metal is a product of the cell reaction.

Explanation:

In writing the line notation of a cell reaction, we ought to recall that the anode is placed at the left hand side of the notation while the cathode is placed at the right hand side of the notation.

If we look at the line notation shown in the question, we will realize that the Cu2+ (aq) | Cu(s) half cell is the cathode and that copper metal is the product of the reaction as shown by the notation.

Answer:

7.15g/cm³

Explanation:

To solve this question we must know that a body-centered cubic unit cell contains 2 atoms.

The volume of the unit cell is:

289pm = (289x10⁻¹²m)³ =

2.414x10⁻²⁹m³ * (1cm³ / 1x10⁻⁶m³) = 2.414x10⁻²³cm³

And the mass is -Molar mass Chromium = 51.9961g/mol:

2atoms * (1mol / 6.022x10²³atoms) * (51.9961g / mol) =

1.727x10⁻²²g

The density is:

1.727x10⁻²²g / 2.414x10⁻²³cm³ =

Answer:

Gravity

Explanation:

All matter exerts a force, which is called , that pulls all other matter towards its center. The strength of the force depends on the mass of the object: the Sun has more gravity than Earth, which in turn has more gravity than an apple. Also, the force weakens with distance. Objects far from the Sun won’t be influenced by its gravity.

Answer:

Sample A

Explanation:

Molarity of a solution can be calculated using the formula:

Molarity = number of moles (mol) ÷ volume (vol)

For Sample A:

V = 300ml = 300/1000 = 0.3 L

Molarity = 1M

n = number of moles (mol)

1 = n/0.3

n = 0.3moles

For Sample B:

V = 145 mL = 145/1000 = 0.145L

Molarity = 1.5 M

n = number of moles

1.5 = n/0.145

n = 1.5 × 0.145

n = 0.22 moles

Based on the above results (moles), sample A with 0.3 moles contains the larger concentration of sodium chloride.

Answer:

b

Explanation:

Answer:

What about it? I don't get the question

Unnillium (101)

Unnilbium (102)

Unniltrium (103)

Unnilquadium (104)

Unnilpentium (105)

Unnilhexium (106)

Unnilseptium (107)

Unniloctium (108)

Unnilennium (109)

Ununnillium (110)

Answer:

weak acid

Explanation:

Answer: Basic is correct!!!!!

Explanation:

Answer:

a.option is the correct answer

Answer:

B.Anything that takes up space and has mass

Answer:

a. i. 8.447 × 10⁻³ T ii.  27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

a.

i. What strength of magnetic field is required?

Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

F = F'

Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

So,

B = mv/rq

B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)

B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

B = 0.8447 × 10⁻² kg/sC)

B = 8.447 × 10⁻³ T

(ii) What is the diameter of the 13C semicircle?

Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and  = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)

r' = 1.357 × 10⁻¹ kgm/TC)

r' = 0.1357 m

r' = 13.57 cm

Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm

b.

i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?

Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

Answer:

meeting ones legal responsibility

Answer:

Explanation:

Given Information :

Mass = 4.0kg

Acceleration  due to gravity = 10 m/s

Weight= ?

[tex]Weight = \frac{mass}{acceleration\: due \:to \:gravity} \\\\W= \frac{4.0}{10} \\\\W= \frac{2}{5} \\\\W=0.4[/tex]

Answer:

50 g Sucrose

Explanation:

Step 1: Given data

Step 2: Calculate the mass of sucrose needed to prepare the solution

The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:

2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose

Answer:

until the next harvest, and seed must be held for the next season's ... successful grain storage is the moisture content of the crop. ... or both. If ambient temperatures are low, then air alone may cool the ... allow some of the drying to take place naturally while the crop ... employed to cool grain that has been placed in storage.

Answer:

Q = 797.5 cal

Explanation:

Given that,

Mass of iron, m = 50 g

The temperature rises from 55°C to 200°C.

We need to find the heat needed to raise the temperature. The heat raised is given by :

[tex]Q=mc\Delta T[/tex]

Put all the values,

[tex]Q=50\times 0.11\times (200-55)\\\\Q=797.5\ cal[/tex]

So, 797.5 calories of heat is needed.

Answer: B

Explanation:

According to Charle's law, at constant pressure the volume of a fixed mass of a ga is directly proportional to its absolute temperature.

At constant pressure, V∝T.

Answer:

The concentration of lead ion required to just initiate precipitation is -[tex]2.37\times10^-^5 M[/tex]

Explanation:

Lets calculate -:

Solubility equilibrium -: [tex]PbI_2(s)[/tex] ⇄ [tex]Pb^2^+ (aq) + 2I^- (aq)[/tex]

Solubility product of [tex]PbI_2[/tex] ,[tex]Q=[Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]=9.8\times10^-^9[/tex]

Concentration of [tex]I^-[/tex][tex]=[KI]=0.0492M[/tex]

When the ionic product exceeds the solubility product , precipitation of salt takes place .

                                   [tex]Q_s_p\geq K_s_p[/tex]

        [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]\geq 9.8\times10^-^9[/tex]

      [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex]  [tex][0.0492]^2[/tex] [tex]\geq 9.8\times10^-^9[/tex]

                       [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq \frac{9.8\times10^-^9}{[0.0492]^2}[/tex]

                        [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]\frac{9.8\times10^-^9}{2.42\times10^-^3}[/tex]

                        [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]2.37\times10^-^5 M[/tex]

Thus , [tex]PbI_2[/tex] will start precipitating when [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex]   [tex]\geq 2.37\times10^-^5 M[/tex].    

Answer:

(C) It is favorable and is driven by ΔH° only.

Explanation:

Let's consider the following balanced equation.

2 Na₂O₂ + S + 2 H₂O → 4 NaOH + SO₂

To determine whether it will be favorable or not at 298 K, we need to calculate the standard Gibbs free energy (ΔG°).

We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T.ΔS°

As we can see from the expression above, the favorability will be driven if ΔH° < 0 and if ΔS° < 0. Since ΔH° = -610 kJ/mol and ΔS° = -73 J/K.mol, the favorability will be driven by ΔH°. Now, let's calculate the overall favorability.

ΔG° = ΔH° - T.ΔS°

ΔG° = -610 kJ/mol - 298 K.(-0.073 kJ/K.mol) = -588 kJ/mol

The reaction is favorable and is driven by ΔH° only.

Answer:

The answer is:

[tex]PART \ A: CH_4\\\\PART \ B: CH_30H\\\\PART \ C: H_2CO[/tex]

Explanation:

Parts A:

The vapor pressure is higher in [tex]CH_4[/tex] because it is non-polar, while [tex]CH_3Cl[/tex] is polar. [tex]CH_4[/tex] has a lower molar weight as well.

Part B:

Although hydrogen bonding is found in both commodities, the vapor pressure is higher because of the smaller molar mass of [tex]CH_30H[/tex].

Part C:

[tex]H_2CO[/tex] does not show hydrogen due to the increased vapor pressure

[tex]CH_3OH[/tex] bonding.

Answer:

P = 487 Torr

mN₂ = 0.347 g

mO₂ = 0.174 g

mHe = 0.0333 g

Explanation:

Step 1: Calculate the total pressure of the mixture

The total pressure of the mixture (P) is equal to the sum of the partial pressures of the gases.

P = pN₂ + pO₂ + pHe

P = 231 Torr + 101 Torr + 155 Torr = 487 Torr

Step 2: Calculate the moles of each gas

We have 3 gases in a 1.00 L container at 25.0 °C (298.2 K). We can calculate the number of moles using the ideal gas equation: P × V = n × R × T.

nN₂ = pN₂ × V / R × T

nN₂ = 231 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.0124 mol

nO₂ = pO₂ × V / R × T

nO₂ = 101 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00543 mol

nHe = pHe × V / R × T

nHe = 155 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00833 mol

Step 3: Calculate the mass of each gas

The molar mass of N₂ is 28.01 g/mol.

mN₂ = 0.0124 mol × 28.01 g/mol = 0.347 g

The molar mass of O₂ is 32.00 g/mol.

mO₂ = 0.00543 mol × 32.00 g/mol = 0.174 g

The molar mass of He is 4.00 g/mol.

mHe = 0.00833 mol × 4.00 g/mol = 0.0333 g

Answer:

Q = -535 J.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the lost energy according to the following and generic heat equation:

[tex]Q=mC(T_F-T_i)[/tex]

Thus, since the specific heat of iron is 0.450 in the SI units, we can plug in the mass and temperatures to obtain:

[tex]Q=28.3g*0.450\frac{J}{g\°C} (24.0\°C-66.0\°C)\\\\Q=-535J[/tex]

Regards

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: [tex]V_{1}[/tex] = 571 mL,       [tex]T_{1} = 26^{o}C[/tex]

(a) [tex]T_{2} = 5^{o}C[/tex]

The new volume is calculated as follows.

[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL[/tex]

(b) [tex]T_{2} = 95^{o}F[/tex]

Convert degree Fahrenheit into degree Cesius as follows.

[tex](1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C[/tex]

The new volume is calculated as follows.

[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL[/tex]

(c) [tex]T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C[/tex]

The new volume is calculated as follows.

[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL[/tex]