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SVEN [57.7K]
3 years ago
15

This is for anyone who’s good with science coz I absolutely hate it. ILL BRAINLIST U AND GIVE U 5 STAR TO WHOEVER IS GOOD;) be s

pecific and use ur own words please! thankyou. and good luck LOL.
conclusions:

The burning candle is an example of a combustion reaction.
a) What do all combustion reactions require?
b) What do all combustion reactions produce?
c) Write a general word equation for the combustion reaction.
Chemistry
2 answers:
marin [14]3 years ago
6 0

Answer:

a) oxygen is nessasary for all combustion to occur

b)carbon dioxide and water

c)for example combustion of methane:

<h3>CH4+2O2 -------> CO2+H2O</h3>
marusya05 [52]3 years ago
4 0

Answer:

A) Oxygen

B) Heat

C)

Explanation:

Hope it helps

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sergejj [24]

Answer: D

Explanation:

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3 years ago
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You are making lemonade and the calls for 6 cups Lemon juice (L), 3 Cups of sugar (S) and 5 cups of water (W) to make 12 Cups of
vova2212 [387]

Answer:

18 Cups

Submitted the assignment

Explanation:

4 0
2 years ago
If an object has a mass of 180 g and a density of 6 g/cm^3, what is the volume of the object?
cupoosta [38]

Answer:

<h3>The answer is 30 cm³</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 180 g

density = 6 g/cm³

We have

volume =  \frac{180}{6}  \\

We have the final answer as

<h3>30 cm³</h3>

Hope this helps you

5 0
3 years ago
How many grams of potassium sulfate should be weighed to prepare 100 ppm of potassium in 250mL of solution if the purity of the
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R=0.0564g K2SO4

4 0
3 years ago
A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system
AleksAgata [21]
To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

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Clearing the specific heat of the unknown substance:

c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}

c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}
6 0
3 years ago
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