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3 years ago

What do plants need to live?

Chemistry

2 answers:

zimovet [89]3 years ago

4 0

Sunlight, soil, water, leaves,

Tpy6a [65]3 years ago

3 0

Answer:

Explanation:

sunlight

soil

water

sugar

carbon dioxide

minerals

leaves

vitamins

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Go back to the simulation and increase the number of molecules of heavy species to 700.

Alekssandra [29.7K]

Answer:

As molecules are added the container inflates/gets larger. ... And the molecules move slower and add pressure.

Explanation:

7 0

3 years ago

Dmitrij [34]

C I’m pretty sure that is the answer

5 0

2 years ago

9. How many moles are in 6.022 x 1024 atoms of oxygen?

ra1l [238]

Answer:

1 mole

Explanation:

The mole (mol) is the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. One mole (1 mol) contains 6.022x1023 entities (to four significant figures).

8 0

2 years ago

What is the molar mass of SO4?

VLD [36.1K]

Yo sup??

the answer is option A ie

96.056 grams/mole

because mass of S is 32 gm and mass of O is 16 gm

Hope this helps

3 0

3 years ago

A scientist measures the standard enthalpy change for the following reaction to be 81.1 kJ :2CO2(g) + 5 H2(g)C2H2(g) + 4 H2O(g)B

posledela

<u>Answer:</u> The enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.Z

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

For the given chemical reaction:

$2CO_2(g)+5H_2(g)\rightarrow C_2H_2(g)+4H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=226.7kJ/mol\\\Delta H^o_{rxn}=81.1kJ$

Putting values in above equation, we get:

$81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol$

Hence, the enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.

8 0

3 years ago