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3 years ago

In the equation C H4 + O2 C O2 + H2 O, what are the products?

Chemistry

2 answers:

Hoochie [10]3 years ago

3 0

The answer to this question is B; the products of this reaction is on the right after after the arrow

Ghella [55]3 years ago

3 0

Answer:

Option B clearly shows the Products of the reaction

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What element has these characteristics MgX2, Very low density, Highly reactive, Diatomic

Kruka [31]

Copper has all those characteristics and even works on a Notatem scale.

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3 years ago

Ome fire extinguishers contain liquid carbon dioxide Suggest how carbon dioxide

melisa1 [442]

Answer:

The CO2 Extinguisher Cannisters contain carbon dioxide in liquid form, and when the extinguisher is let off the liquid is released into the air neutralising the oxygen that the fire is feeding on, disabling the fires ability to spread.

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3 years ago

Penicillinase, also known as β‑lactamase, is a bacterial enzyme that hydrolyzes and inactivates the antibiotic penicillin. Penic

sergeinik [125]

Explanation:

The given data is as follows.

$V_{max} = 6.8 \times 10^{-10} \mu mol/min$

$K_{m} = 5.2 \times 10^{-6} M$

Now, according to Michaelis-Menten kinetics,

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

where, S = substrate concentration = $10.4 \times 10^{-6}$ M

Now, putting the given values into the above formula as follows.

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667$

= $4.5 \times 10^{-10} \mu mol/min$

This means that $V_{o}$ would approache $V_{max}$ .

5 0

4 years ago

A student measures the mass of an 8 cm^3 block of brown sugar to be 12.9 g. What is the density of

Oduvanchick [21]

Answer:

1.6125 g/cm^3

Explanation:

Density= mass/volume

4 0

3 years ago

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add p

Butoxors [25]

Answer:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Explanation:

Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

$BiO_3^-\rightarrow Bi^{3+}$

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

$(Bi^{5+}O_3)^-\rightarrow Bi^{3+}$

Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

$6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O$

Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

$6OH^-+6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Then, we accommodate the waters to obtain:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Best regards!

7 0

3 years ago