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cluponka [151]
2 years ago
6

Which term should replace the question mark?

Chemistry
1 answer:
ruslelena [56]2 years ago
5 0
Where are the terms ? I don’t know how to replace the question mark with a term if there didn’t any terms to decide from
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How many joules of energy are made by the loss of 100kg of mass using the formula E=mc2? please help!!!
asambeis [7]

Answer:

E = 3 × 10¹⁰ J

Explanation:

Mass, m = 100 kg

We need to find energy made by the loss of 100 kg of mass. The formula between the mass and energy is given by :

E = mc²

Where c is speed of light

Putting all the values, we get :

E = 100 kg × (3×10⁸ m/s)²

= 3 × 10¹⁰ J

So, the required energy is 3 × 10¹⁰ J.

6 0
3 years ago
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AnnZ [28]
Here you are looking on the Free Body diagram of a net force of 0N in both the x and y-directions.  the only ones that has that condition met is A and C.
4 0
3 years ago
Write the isotope notation for an element with 42 protons and 96 neutrons.
tankabanditka [31]

Answer:138

Explanation:

4 0
3 years ago
Read 2 more answers
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A?
BartSMP [9]
The mass of gas A is 11.56. i got this answer by 0.68 multiplied by 17.  because in this problem the key word is times.
4 0
3 years ago
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