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Leno4ka [110]
1 year ago
9

Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.

Chemistry
1 answer:
pogonyaev1 year ago
7 0
I think it's balanced ?
There are two H's in the both side
And one P in the both side
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Which element has an atomic number of 16?<br> a. oxygen (O)<br> b. sulfur (S)<br> c. germanium (Ge)
emmasim [6.3K]

Answer: Sulfur

Explanation:

3 0
3 years ago
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A certain orbital of the hydrogen atom has n = 4 and l = 2 what are the possible values of ml for this orbital?
Nadya [2.5K]

The ml is also called as the magnetic quantum number. The value of ml can range from –l to +l including zero. Hence all of the possible values for ml given that l = 2 are:

<span>-2, -1, 0, + 1, + 2</span>

3 0
3 years ago
Please Help!
Whitepunk [10]

We can use two equations for this problem.<span>

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is decay constant.

20 days = 0.693 / λ 
λ   = 0.693 / 20 days        (1) 

Nt = Nο eΛ(-λt)                (2)

Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.
t = 40 days</span>

<span>No = 200 g

From (1) and (2),
Nt =  200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>

</span>Hence, 50.01 grams of isotope will remain after 40 days.

<span>
</span>

3 0
3 years ago
What is atom economy?
bonufazy [111]

Answer:

The conversion efficiency of a chemical process.

Explanation:

Hope this helps!

5 0
3 years ago
Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl
tensa zangetsu [6.8K]

Answer:

[SO_2Cl_2] = 0.09983 M

Explanation:

Write the balance chemical equation ,

SO_2Cl_2((g) = SO_2(g) + Cl_2(g)

initial concenration of SO_2Cl_2((g)  =0.1M

lets assume that degree of dissociation=\alpha

concenration of each component at equilibrium:

[SO_2Cl_2] = 0.1-0.1\alpha

[SO_2] = 0.1\alpha

[Cl_2] = 0.1\alpha

Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}

Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}

as \alpha is very small then we can neglect  1-\alpha

therefore ,

Kc ={0.1\alpha \times \alpha}

\alpha =\sqrt{\frac{Kc}{0.1}}

\alpha = 1.73 \times 10^{-3}

Eqilibrium concenration of [SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173

[SO_2Cl_2] = 0.09983 M

4 0
3 years ago
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