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My name is Ann [436]
3 years ago
13

How many grams of carbon dioxide are produced when 191 g of methane (CH4) are burned completely as shown in the reaction below:

Chemistry
1 answer:
guajiro [1.7K]3 years ago
4 0
CH5
step-by-step-explanation
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How is the life cycle of a star similar to that of a human?
QveST [7]
Stars have a life cycle, just like people: they are born, grow, change over time, and eventually grow old and die. Most stars change in size, color, and class at least once in their lifetime.
Brainliest?
4 0
2 years ago
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(e) Another student investigated the rate of a different reaction
vodka [1.7K]

Answer:

0.07 g/s.

Explanation:

From the question given above, the following data were obtained:

Mass lost = 9.85 g

Time taken = 2 min 30 s

Mean rate =?

Next, we shall convert 2 min 30 s to seconds (s). This can be obtained as follow:

1 min = 60 s

Thus,

2 min = 2 × 60 = 120 s

Therefore,

2 min 30 s = 120 s + 30 s = 150 s

Finally, we shall determine the mean rate of the reaction. This can be obtained as illustrated below:

Mass lost = 9.85 g

Time taken = 150 s

Mean rate =?

Mean rate = mass lost / time taken

Mean rate = 9.85 / 150

Mean rate = 0.07 g/s

Therefore, the mean rate of the reaction is 0.07 g/s

4 0
3 years ago
if a plot weight (in g) vs. volume (in ml) for a metal gave the equation y= 13.41x and r^2=0.9981 what is the density of the met
Bumek [7]

ANS: density = 13.41 g/ml

Density (d) of a substance is the mass (m) occupied by it in a given volume (v).

Density = mass/volume

i.e. d = m/v

m = (d) v -----(1)

The given equation from the plot of weight vs volume is :

y = 13.41 x ----(2)

Based on equations (1) and (2) we can deduce that the density of the metal is 13.41 g/ml

8 0
3 years ago
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How do you think winds affect air pollution?
Elza [17]
Wind affects pollution because it moves it, the wind carries the pollution and move it somewhere else
7 0
3 years ago
A sample of gas has a volume of 20.0 mL at STP. What will the volume be if the temperature is changed to 546 K and the pressure
Ostrovityanka [42]

The volume did not change, it remained at 20 ml

<h3>Further explanation</h3>

Given

20 ml a sample gas at STP(273 K, 1 atm)

T₂=546 K

P₂=2 atm

Required

The volume

Solution

Combined gas Law :

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

Input the value :

\tt \dfrac{1\times 20}{273}=\dfrac{2\times V_2}{546}\\\\V_2=\dfrac{1\times 20\times 546}{273\times 2}\\\\V_2=20~ml

The volume does not change because the pressure and temperature are increased by the same ratio as the initial conditions (to 2x)

5 0
3 years ago
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