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3 years ago

14 points!!! Pls tell me answer :3

Chemistry

2 answers:

iren [92.7K]3 years ago

8 0

I’d say the last one

trasher [3.6K]3 years ago

7 0

It's either the last or the first one. I think the first one makes more sense though because it has already told you it created offspring so saying that again seem kind of redundant. Sorry I'm not sure

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SiCl4(l) + H2O(l) → SiO2(s) + HCl(aq)

Sveta_85 [38]

I assume you’re looking for a balanced equation.
SiCl4 + 2H2O = SiO2 + 4HCl

5 0

4 years ago

IDENTIFY THE NUMBER OF SIGNIFICANT FIGURES IN EACH OF THESE MEASUREMENTS OF AN OBJECT: 76.47 AND 76.59

murzikaleks [220]

Answer:

they both have 4 sig figs:)

Explanation:

8 0

3 years ago

11. Why is HCl a strong acid and HClO a weak acid?

ankoles [38]

Strong acid:dissolves and dissociates 1005 to produce protons (H+) 1. seven
strong acids: HCI, HBr, HI, HNO3, H2SO4, and HCIO3. ...
weak acid: dissolves but less than 100% dissociates to produce protons (H+) 1.

3 0

3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

3 0

3 years ago

If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K and then I raise the pressu

mixas84 [53]

Answer:

The answer to your question is V2 = 29.6 l

Explanation:

Data

Pressure 1 = P1 = 12 atm

Volume 1 = V1 = 23 l

Temperature 1 = T1 = 200 °K

Pressure 2 = 14 atm

Volume 2 = V2 = =

Temperature 2 = T2 = 300°K

Process

1.- To solve this problem use the Combine gas law.

P1V1/T1 = P2V2/T2

-Solve for V2

V2 = P1V1T2 / T1P2

2.- Substitution

V2 = (12)(23)(300) / (200)(14)

3.- Simplification

V2 = 82800 / 2800

4.- Result

V2 = 29.6 l

5 0

3 years ago