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3 years ago

13

What is the name of the scientist who discovered the atoms have positive charges

2 answers:

Alex Ar [27]3 years ago

7 0

It was Rutherford who discovered this

Setler [38]3 years ago

5 0

<span>It was Rutherford who discovered that the positive charge part of the atom.
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GUYS I REALLY NEED THIS ASAP!!!!!

Nastasia [14]

Answer:

I don't think so

Explanation:

The equation doesn't look balanced

8 0

3 years ago

What is the mass of 6 atom(s) of copper in grams?

USPshnik [31]

Answer:

6.33×10¯²² g

Explanation:

From the question given above, the following data were obtained:

Number of atoms = 6 atoms

Mass of copper (Cu) =?

From Avogadro's hypothesis, we understood that:

6.02×10²³ atoms = 1 mole of Cu

But 1 mole of Cu = 63.5 g

Thus,

6.02×10²³ atoms = 63.5 g of Cu

Finally, we shall determine the mass of 6 atoms of copper. This can be obtained as illustrated below:

6.02×10²³ atoms = 63.5 g of Cu

Therefore,

6 atoms = (6 × 63.5) / 6.02×10²³

6 atoms = 6.33×10¯²² g of Cu

Therefore, the mass of 6 atoms of copper is 6.33×10¯²² g.

4 0

3 years ago

What is the charge on an atom with an equal number of protons and electrons?

dexar [7]

Answer: the answer is a netrual bond

Explanation: a netural bond happens when the number of electrons are the same as the number of protons.

7 0

3 years ago

Read 2 more answers

Which of these compounds is a product of protein synthesis?

yuradex [85]

The answer is c it is Carbon Dioxide

3 0

3 years ago

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago