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Sindrei [870]
3 years ago
6

Where is the most of the mass of an atom located?

Chemistry
1 answer:
kow [346]3 years ago
4 0
<span>The mass of an atom located in the A) nucleus.</span>
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Why does water have a lopsided shape and act like a chemical magnet?
leva [86]

Answer:

See below  

Step-by-step explanation:

(a) Shape

The formula for water is H-O-H.

The central O atom has four electron pairs around it. They try to get as far from each other as possible, so they point toward the corners of a tetrahedron.

Only two of the pairs have a hydrogen atom attached, so water has a bent shape. The H-O-H bond angle is about 104°.

(b) Chemical magnet

The O atom has a greater attraction than H for the shared electrons in the O-H bonds, so the electrons spend more time near the O.

This gives the O atom a partial negative charge (pink in the diagram) and the H atoms a partial positive charge (blue).

The water molecule acts like a chemical magnet because its negative end attracts the positive ends of other molecules, while its positive ends attract the negative ends of other molecules.

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3 years ago
An electrode reaction is reduction. This means that the electrode is a(n) a. cathode. c. electrochemical cell. b. anode. d. elec
Romashka [77]
Oxidation happens at the anode and reduction happens at the cathode.<span />
5 0
3 years ago
How well can you apply Charles’s law to this sample of gas that experiences changes in pressure and volume? Assume that pressure
anzhelika [568]

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A: 384

B: 0.85

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3 0
3 years ago
Read 2 more answers
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
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