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I am Lyosha [343]
3 years ago
6

Check all that apply to sugar.

Chemistry
2 answers:
Sav [38]3 years ago
6 0

Answer:

Does not conduct electricity and is a non-electrolyte

Explanation:

Because I pay attention

harina [27]3 years ago
3 0

Answer:

Candy

Explanation:

its sweet and has a lot of sugar and acid

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Which of the following is true about the frequencies of electromagnetic radiation
Semmy [17]
<h2>Answer:Ultraviolet waves have higher frequencies than radio waves.</h2>

Explanation:

The waves arranged in increasing order of frequency are radio waves,infrared waves,x-rays,ultraviolet waves.

Option A:

Ultraviolet waves have higher frequencies than radio waves.This agrees with the above order.

Option B:

X-rays have lower frequencies than infrared waves.This does not agree with the above order.

Option C:

Radio waves have higher frequency than x-rays.This does not agree with the above order.

Option D:

Gamma rays have lower frequencies than ultraviolet waves.This does not agree with the above order.

So,option A is correct.

3 0
4 years ago
The elementary gas-phase reaction2A → Bis carried out in a constant-volume batch reactor where 50% conversion is achieved in 1 h
Zina [86]

Answer:

V=5000dm^3

Explanation:

Hello,

In this case, for the STR reactor carrying out this chemical reaction, the material balance is:

\frac{dC_A}{dt}=r_A\\\\\frac{dC_A}{dt}=-kC_A^2

Thus, with the given information we compute the rate constant as follows:

C_A_0 \frac{dX_A}{dt}=kC_A_0^2(1-X_A)^2\\\\\int\limits^{0.5}_0 {\frac{dX_A}{(1-X_A)^2} } \,=kC_A_0t\\\\\frac{0.5}{1-0.5}=k*0.2\frac{mol}{dm^3}*1h\\ \\k=\frac{1}{(0.2\frac{mol}{dm^3})*1h} \\\\k=5\frac{dm^3}{mol*h}

Now, for the CSTR we have the following design equation in terms of conversion for finding the volume:

V=\frac{F_A_0*X_A}{-r_A}\\\\V=\frac{F_A_0*X_A}{k*C_A_0^2(1-X_A)^2}

Therefore, it turns out:

V=\frac{500\frac{mol}{h} *0.5}{5\frac{dm^3}{mol*h} *(0.2\frac{mol}{dm^3} )^2(1-0.5)^2}\\\\V=5000dm^3

Best regards.

7 0
3 years ago
Pt.3 of my questions because a bot answer it and no one could answer it
Andrei [34K]

Answer:

I cant see the whole question but to my knowledge it is the 3rd law

Explanation:

because the third law states what the applied force is when two objects interact

4 0
3 years ago
Suppose that 0.25 mole of gas C was added to the mixture without changing the total pressure of the mixture. How does the additi
Yanka [14]

Here we have to get the effect of addition of 0.25 moles of gas C on the mole fraction of gas A in a mixture of gas having constant pressure.

On addition of 0.25 moles of C gas, the mole fraction of gas A will be  \frac{moles of gas A}{moles of gas A + 0.25}.

The partial pressure of gas A can be written as P_{A} = x_{A}×P (where x_{A} is the mole fraction of gas A present in the mixture and P is the total pressure of the mixture.

The mole fraction of gas A in a mixture of gas A and C is = \frac{moles of gas A}{moles of gas A + moles of gas C} and \frac{moles of gas C}{moles of gas A + moles of gas C} respectively.

Thus on addition of 0.25 moles of C gas, the mole fraction of gas A will be  \frac{moles of gas A}{moles of gas A + 0.25}.

Which is different from the initial state.

4 0
3 years ago
Read 2 more answers
In a solution of water and table salt, what is the table salt?
g100num [7]

the salt is the solute (and the water is the solvent) i hope this answers your question lol

6 0
4 years ago
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