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3 years ago

A high level of ________________ in your urine could show that you have diabetes.

Chemistry

1 answer:

Ksenya-84 [330]3 years ago

7 0

Answer:

ketones.

Explanation:

If you have a high level of ketones in your urine, it usually means you're not getting enough insulin, which is a sign of Diabetes.

(I got this Information from the "Ketones in Urine" article from MedlinePlus.)

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2567 milliliters to liters

ollegr [7]

Answer:

2.567

Explanation:

To change milliliters into liters you divide by 1000

So when you divide 2567 by 1000 you get 2.567

6 0

3 years ago

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Give two significant consequences of the Copernican Revolution!! PLEASE HELP! NEED ANSWER NOW!

Marina CMI [18]

Hi !
1.the theological perspective of man's dominance in a dominant world was shown to be no longer tenable.
2. Heliocentricity uprooted the theological doctrine that the universe revolved around Man as the paramount and most glorious end-result of Creation, with the Sun, the Stars, the other Planets and the whole Universe .

Hood it helps and have a wonderful day!

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4 years ago

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The Sierra Nevada, a mountain range in California, is made

tensa zangetsu [6.8K]

Id say uplift but i’m not sure..

7 0

3 years ago

What should I do to order to get the answer?

jarptica [38.1K]

Answer:

Try looking back at what to do and reading back through it to understand it.

Explanation:

5 0

3 years ago

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

4 0

3 years ago