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Andre45 [30]
2 years ago
14

What kind of weather does a cold front usually bring?

Chemistry
2 answers:
svlad2 [7]2 years ago
7 0

Answer:

Sunny and windy but mostly windy

aleksandr82 [10.1K]2 years ago
7 0
I believe the answer is C
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Which of the following would contain 48 electrons?
Helga [31]
The answer is cadmium its got 48 electrons which is y its number 48 on the period table
4 0
2 years ago
Solve for b if R = x(A + B)
Tatiana [17]

This is what I got. Hope it helps :)

7 0
3 years ago
6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets
Nata [24]

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: V_{1} = 0.20 mL,         T_{1} = 27^{o}C

V_{2} = ?,                    T_{2} = 177^{o}C

Formula used to calculate new volume is as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL

Thus, we can conclude that the new volume of cake is 1.31 mL.

5 0
3 years ago
What is the binding energy of a nucleus that has a mass defect of 5.81*10-^29 kg
IrinaVladis [17]

Answer:

Choice A: Approximately 5.23 \times 10^{-29} joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the \rm 5.81\times 10^{-29} kilograms of mass.

E = m \cdot c^{2},

where

  • E stands for energy,
  • m stands for mass, and
  • c is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of c?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}.

Apply the mass-energy equivalence:

\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}.

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

\rm 1\; J = 1\; N \times 1\; m.

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}.

In other words, the mass defect here is also \rm 5.23\times 10^{-12}\; J.

7 0
3 years ago
Read 2 more answers
Exercise A mixture of 250 mL of methane, CH4, at 35 °C and 0.55 atm and 750 mL of propane, C3Hg, at 35° C and 1.5 atm, were intr
dalvyx [7]

Explanation:

The given data is as follows.

      V_{1} = 250 mL,     V_{2} = 750 mL

      T_{1} = 35^{o}C = 35 + 273 K = 308 K

      T_{2} = 35 + 273 K = 308 K

      P_{1} = 0.55 atm,    P_{2} = 1.5 atm

               P = ? ,         V = 10.0 L

Since, temperature is constant.

So,    P_{1}V_{1} + P_{2}V_{2} = PV

Now, putting the given values into the above formula as follows.

         P_{1}V_{1} + P_{2}V_{2} = PV

         0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm = P \times 10.0 L

                     P = 0.126 atm

As, 1 atm = 760 torr. So, 0.126 atm \times \frac{760 torr}{1 atm} = 95.76 torr.

Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.

8 0
2 years ago
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