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Olenka [21]
3 years ago
5

1)

Chemistry
2 answers:
dalvyx [7]3 years ago
7 0

All I know for certain is Potential energy is converted to kinetic energy.

That is the answer to question 3.

Andrej [43]3 years ago
4 0
1.<span>B) <span>the subscript
2.<span>B)<span>CO2 </span></span>.
3.B) potential energy is converted into kinetic energy
4.A) A roller coaster at the top of a hill
6.C) in motion
7.D) weed whacker full of gas
8.B) potential energy only
9.</span></span>A) baking soda
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What are the roles of producers,consumers, and decomposers in an ecosystem?
scoundrel [369]
The producers of an ecosystem make energy available to all the other living parts of an ecosystems. Consumers like humans eat the food that producers like plants made decomposers decompose dead plants and animals when they die like bacteria is a decomposer
5 0
3 years ago
Read 2 more answers
Consider water at 500 kPa and a specific volume of 0.2 m3/kg, what is the temperature (in oC)?
Artyom0805 [142]

Answer:

T=-272.9^{o}C

Explanation:

We have the ideal gasses equation PV=nRT and the expression for the specific volume v=\frac{V}{m}, that is the inverse of the density, and for definition the number of moles is equal to the mass over the molar mass, that is n=\frac{m}{M}

And we can relate the three equations as follows:

PV=nRT

Replacing the expression for n, we have:

PV=\frac{m}{M}RT

P\frac{V}{m}=\frac{RT}{M}

Replacing the expression for v, we have:

Pv=\frac{RT}{M}

Now resolving for T, we have:

T=\frac{PvM}{R}

Now, we should convert all the quantities to the same units:

-Convert 500kPa to atm

500kPa*\frac{0.00986923}{1kPa}=4.93atm

-Convert 0.2\frac{m^{3}}{kg} to \frac{L}{kg}

0.2\frac{m^{3} }{kg}*\frac{1L}{1m^{3}}=0.2\frac{L}{kg}

- Convert the molar mass M of the water from \frac{g}{mol} to \frac{kg}{mol}

18\frac{g}{mol}=\frac{1kg}{1000g}=0.018\frac{kg}{mol}

Finally we can replace the values:

T=\frac{(4.93atm)(0.2\frac{L}{kg})(0.018\frac{kg}{mol})}{0.082\frac{atm.L}{mol.K}}

T=0.216K

T=0.216K-273.15\\T=-272.9^{o}C

5 0
3 years ago
If a reaction mixture contains 24 g of Mg and 32 g of O2 , what is the limiting reactant?
Alexxx [7]

This problem is providing the mass of both magnesium metal and oxygen gas and involved in a chemical reaction and asks for the limiting reactant. At the end, it turns out to be identified as magnesium.

<h3>Stoichiometry</h3>

In chemistry, stoichiometry is a widely-used tool we use in order to relate the mass and moles of different chemical substances involved in a chemical reaction. Thus, we consider the following chemical equation between magnesium and oxygen to produce magnesium oxide.

2Mg+O_2\rightarrow 2MgO

However, when the mass of the both of the reactants is given, one must identify the limiting reactant as the one producing the least of the moles of the product, which means we can use the given grams of the both of the reactants, their molar masses and mole ratios with the product to obtain the aforementioned:

24gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}=0.988molMgO\\ \\ 32gO_2*\frac{1molO_2}{32.0gO_2}*\frac{2molMgO}{1molO_2}=2molMgO

Thus, we can evidence how 24 g of magnesium produce the least of the moles of magnesium oxide, fact validating the magnesium as the limiting reactant and the oxygen as the excess one.

Learn more about stoichiometry: brainly.com/question/9743981

3 0
2 years ago
How do atomic and molecular interactions explain the properties of matter that we see and feel
Licemer1 [7]

Answer:they change

Explanation:

just because

8 0
3 years ago
Forces that act between two molecules are referred to as intermolecular forces true or false
julsineya [31]
Answer:
            TRUE: <span>Forces that act between two molecules are referred to as Intermolecular Forces.

Explanation:
                   Those forces which are present within the molecule among atoms are called as Intramolecular Forces, while, The forces which are present between two molecules are called as Intermolecular Forces. Intermolecular Forces are as follow,

                1) Hydrogen Bond Interactions

                2) Dipole-Dipole Interactions

                3) London Dispersion Forces</span>
3 0
3 years ago
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