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2 years ago

A student draws the model shown below. Which of these best compares the conditions at Location X and Location Y?

Chemistry

1 answer:

arlik [135]2 years ago

5 0

Answer:

Explanation:

You have to use formula b to your answer

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The primary forces of attraction between water molecules in H2O(l) are

MakcuM [25]

The answer is Hydrogen bonds

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3 years ago

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Describe some of the solutions that engineers might propose for maximizing the percent yield of hydrogen from discarded wood.

belka [17]

Tamam öğretmenim teşekkür ederiz çok güzel olmuş mu bir

8 0

2 years ago

Will give brainly if correct

nydimaria [60]

Answer:

3.4 mol Li2SO4

Explanation:

6.8 mol LiOH × (1 mol Li2SO4/2 mol LiOH)

= 3.4 mol Li2SO4

8 0

3 years ago

List the three requirements for a correctly written chemical equation.2. Give one example of a word equation, one example of a f

erma4kov [3.2K]

Answer:

Requirements for a correctly written chemical equation are reactants and products, their formula and valency

Explanation:

Formula of the given compound are -

1 - Potassium Hydroxide - $KOH$

2 - Calcium Nitrate - $Ca(NO_3)_2$

The requirements for a correctly written chemical equation are -

Identifying reactants and products
Formula of reactants and products
Valency of elements

Example of word equation, formula equation, and chemical equation is as follows -

Aluminium + iron9(III)oxide ⇒ aluminium oxide + iron (word equation)

$Al_(_s_)$ + $Fe_2O_3_(_s_)$ ⇒ $Al_2O_3_(_s_)$ + $Fe_(s)$ (formula equation)

$2Al_(_s_)$ + $Fe_2O_3_(_s_)$ ⇒ $Al_2O_3_(_s_)$ + $2Fe_(_s_)$ (chemical equation)

8 0

3 years ago

Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl

anzhelika [568]

Answer: ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

$\Delta H=-277.4kJ$

Therefore, the enthalpy change for this reaction is, -277.4 kJ

4 0

3 years ago