What is the first anmial

Which of the following statements about the combustion of glucose with oxygen to form water and carbon dioxide (C6H12O6 + 6 O2 →

KATRIN_1 [288]

Answer:

what are your statements?

3 0

3 years ago

Which of the following gases experience the greatest degree of deviation from ideal behavior under high pressures and low temper

xz_007 [3.2K]

A) CH4

In general, methane reactions are difficult to control. Partial oxidation to methanol, for example, is a rather difficult reaction because the chemical reactions that occur continue to form carbon dioxide and water even though the amount of oxygen available is insufficient.

<h2>Further explanation </h2>

Methane is the simplest hydrocarbon in the form of gas with the chemical formula CH4. Pure methane does not smell, but if used for commercial purposes, a bit of sulfur is usually added to detect leaks that might occur.

Methane is a greenhouse gas. Methane is used in chemical industrial processes and can be transported as frozen liquids (liquefied natural gas, or LNG).

Methane is a major component of natural gas, around 87% of volume.

Methane is not toxic, but is highly flammable and can cause explosions when mixed with air.

Learn More

CH4 / Methane brainly.com/question/9473007

Benefits of methane brainly.com/question/10818009

Details

Class: college

Subject: chemistry

Keywords: ch4, methane, chemicals

8 0

3 years ago

Fermium-253 has a half-life of 0.334 seconds. How long will it take, in seconds, for a 96 atom sample to decay to 3 atoms?

iris [78.8K]

Answer:3.34 seconds

Explanation:

8 0

3 years ago

The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the

IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

Explanation :

The expression for first order reaction is:

$[C_t]=[C_o]e^{-kt}$

where,

$[C_t]$ = concentration at time 't' (final) = ?

$[C_o]$ = concentration at time '0' (initial) = 0.100 M

k = rate constant = $5.40\times 10^{-3}s^{-1}$

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

$[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}$

$[C_t]=4.05\times 10^{-4}M$

Thus, the concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

4 0

3 years ago

A solution contains 0.115 mol h2o and an unknown number of moles of nacl. the vapor pressure of the solution at 30°c is 25.7 tor

Valentin [98]

According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m

8 0

3 years ago