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4 years ago

What does the electron sea model for metals suggest?

Chemistry

1 answer:

fgiga [73]4 years ago

8 0

Answer:

electron sea model for metals suggest that valence electrons drift freely around the metal cations.

Explanation:

Explanation: In electron sea model, the valence electrons in metals are delocalized instead of orbiting around the nucleus. ... These electrons are free to move within the metal atoms. Thus, we can conclude that the electron sea model for metals suggest that valence electrons drift freely around the metal cations.

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In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)

lyudmila [28]

Answer : The mass of $NaBr$ is, 295.323 grams

Solution :

First we have to calculate the moles of $NaNO_3$ .

$\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles$

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaNO_3$ react with 2 moles of $NaBr$

So, 2.87 moles of $NaNO_3$ react with 2.87 moles of $NaBr$

Now we have to calculate the mass of NaBr.

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

$\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g$

Therefore, the mass mass of $NaBr$ is, 295.323 grams

7 0

3 years ago

If a student mixes 75 mL of 1.30 M HNO3 and 150 mL of 6.5 M NaOH. is the final solution acidic, basic, or neutral

raketka [301]

Answer:

The solution is basic.

Explanation:

We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):

If no. of millimoles of acid > that of base; the solution is acidic.

If no. of millimoles of acid = that of base; the solution is neutral.

If no. of millimoles of acid < that of base; the solution is basic.

We need to calculate the no. of millimoles of acid and base:

no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.

no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.

<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>

<em>So, the solution is: basic.</em>

4 0

3 years ago

Please answer fast, shouldn't take long to do ^-^ (will give brainliest)

Anna [14]

Answer:

Explanation:

5 0

3 years ago

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ira [324]

Answer:

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2 years ago

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shepuryov [24]

Answer:

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3 0

2 years ago