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3 years ago

Help me pleaseeeeeeee

Chemistry

2 answers:

Goshia [24]3 years ago

8 0

Answer:

hope it's helpful ❤❤❤❤❤❤

THANK YOU.

anzhelika [568]3 years ago

4 0

Answer:

Explanation:

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Which element will have the greatest attraction for the calcium electrons?

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Its would be fluorine as its the most electronegative

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3 years ago

How many liters of

Kruka [31]

The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂

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3 years ago

Convert 35 g of iron to moles of iron.

Ad libitum [116K]

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2 years ago

Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha

DochEvi [55]

Answer: When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

$CaO(s)+CO_2\rightarrow CaCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)$ $\Delta H_1=-812.8kJ$

(2) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$ $\Delta H_2=-1269kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[\frac{1}{2}\times (-\Delta H_2)]$

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

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