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3 years ago

H2+o2 - h2o mass Please help

Physics

1 answer:

emmasim [6.3K]3 years ago

4 0

H2O2 H2O

H2 + O2 → H2O. Word equation: Hydrogen gas + Oxygen gas → water. Type of Chemical Reaction: For this reaction we have a Combination reaction. Balancing Strategies: For this reaction it is helpful to start by changing the coefficient in front of H2O and so that you have an even number of oxygen atoms.

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A boy is pedaling his bicycle at a velocity of 0.20km/ minute . How far will he travel in 2.5 hours

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30km. 24 the first two hours and 6 the half hour

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3 years ago

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The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

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3 0

3 years ago

Look at the graph

viva [34]

Here, Carefully look at the graph.
When it is on x=10, it is approximately 10, (slightly less than 10)
Closest value would be 90, so y/x = 90/10 = 9

So, the density of the graph would be 9 g/cm³

In short, Your Answer would be Option D

Hope this helps!

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3 years ago

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Answer:

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Explanation:

GIVEN DATA:

$\gamma = 5/3$