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jeka94
3 years ago
11

Is kelp a living or non-living thing?

Chemistry
1 answer:
Fed [463]3 years ago
3 0

Answer:

non living

Explanation:

they trying to trick you

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
Anni [7]

Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

7 0
3 years ago
When a warm air mass and a cold air masses meet but they push at each other and neither one can move (at a stand off) it’s calle
Tom [10]
This is called a Cold Front.
8 0
3 years ago
Which is a characteristic of most nonpolar compounds?
iVinArrow [24]
Nonpolar compound would be symmetrical
4 0
3 years ago
Read 2 more answers
Given that nitrogen forms three bonds with hydrogen to make <img src="https://tex.z-dn.net/?f=NH_%7B3%7D" id="TexFormula1" title
lbvjy [14]

Answer:

Three hydrogen atoms to form PH₃.

Explanation:

Hello!

In this case, since the elements belonging to the nitrogen family (N, P, As, Sb and Bi) show five valence electrons, because there are five electrons at their outer shell, it is clear that if phosphorous bonds with hydrogen, it is going to require the same amount of oxygen atoms (3) because elements having five valence electrons need 3 bonds in order to attain the octet (5+3=8).

Therefore the compound would be:

PH_3

Which is phosphine.

Best regards!

3 0
3 years ago
The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con
bonufazy [111]

<u>Answer:</u> The concentration of radon after the given time is 3.83\times 10^{-30}mol/L

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=3.82days

Putting values in above equation, we get:

k=\frac{0.693}{3.82}=0.181days^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant = 0.181days^{-1}

t = time taken for decay process = 3.00 days

[A_o] = initial amount of the reactant = 1.45\times 10^{-6}mol/L

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}

[A]=3.83\times 10^{-30}mol/L

Hence, the concentration of radon after the given time is 3.83\times 10^{-30}mol/L

7 0
3 years ago
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