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2 years ago

Does the Caribbean have high or low viscosity lava?

Chemistry

1 answer:

butalik [34]2 years ago

8 0

High now I need to write 20 characters

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How much water (H2O ) would form if 4.04 g of hydrogen (H2) reacted with 31.98 g of oxygen (O2 )?

Norma-Jean [14]

Answer:

Mass = 36 g

Explanation:

Given data:

Mass of water formed = ?

Mass of hydrogen = 4.04 g

Mass of oxygen = 31.98 g

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 4.04 g/ 2 g/mol

Number of moles = 2.02 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 31.98 g/ 32 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of water with hydrogen and oxygen.

O₂ : H₂O

1 : 2

H₂ : H₂O

2 : 2

2.02 : 2.02

Number of moles of water formed by oxygen are less thus oxygen will limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 2 mol × 18 g/mol

Mass = 36 g

8 0

2 years ago

1. Water vapor has_________ energy because it is a __________.

natulia [17]

Water vapor has more energy because it is a gas :)

4 0

3 years ago

Ok, I need help with this one!!!!!!!!!!!!!!!!!!!! plz and thank you

Kitty [74]

Answer:

Explanation:

3 0

3 years ago

How many glucose molecules can be produced per second in a plant

Masja [62]

Since glycolysis of one glucose molecule generates two acetyl CoA molecules, the reactions in the glycolytic pathway and citric acid cycle produce six CO2 molecules, 10 NADH molecules, and two FADH2 molecules per glucose molecule

3 0

2 years ago

Consider the reaction 2 NO + O2 → 2 NO2

Sonbull [250]

Answer:

(a) Rate at which $NO_2$ is formed is 0.050 M/s

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

$2NO+O_2\rightarrow 2NO_2$

The expression for rate can be written as:-

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

Given that:- $\frac{d[NO]}{dt}=-0.050\ M/s$ (Negative sign shows consumption)

$-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

$-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}$

$-(-0.050\ M/s)=\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=0.050\ M/s$

(a) Rate at which $NO_2$ is formed is 0.050 M/s

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}$

$-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}$

$\frac{d[O_2]}{dt}=0.0250\ M/s$

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

5 0

2 years ago