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Lorico [155]
4 years ago
14

Nvhfgfghjhlkllll ijdfijdalfjbhhed

Chemistry
1 answer:
Grace [21]4 years ago
8 0
I relate to this so much
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What is the pressure of 2.20 mol of a gas stored in a 4.5 L container at a temperature of 170 K? (Use PV=nRT and R = 8.314 L∙kPa
algol13

Answer:

I don't know how can i do

Explanation:

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7 0
3 years ago
The charge on the most stable ion of zinc​
Natalija [7]
Zinc (Zn) always has a +2 charge. It is one of the exceptions.
8 0
3 years ago
Reactants → products
uranmaximum [27]

Answer:

they are equal

Explanation:

the Law of Conservation of Mass states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system's mass cannot change

3 0
4 years ago
An unknown element sample has 2 isotopes present. The first isotope has a mass of 6.017 amu and is
oksano4ka [1.4K]

Answer:

Calculating Atomic Mass

Change each percent abundance into decimal form by dividing by 100. Multiply this value by the atomic mass of that isotope. Add together for each isotope to get the average atomic mass.

Explanation:

have a nice day

8 0
2 years ago
If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?
myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30.17}\ year^{-1}

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Initial concentration [A_0] = ?

Final concentration [A_t] = 12.5 grams

Applying in the above equation, we get that:-

12.5\ g=[A_0]e^{-0.02297\times 90.6}

[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g

<u>Mass of original sample = 100 g</u>

8 0
3 years ago
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