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3 years ago

The element nickel has the following properties. Select the two that are chemical properties of nickel.

Chemistry

1 answer:

kkurt [141]3 years ago

5 0

a. It melts at 1455oC I know this is correct I need One more

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Chem homework, help?

IrinaVladis [17]

I'm not writing random things to reach the 20 character requirement so I can tell you that Magnesium has the least amount of atoms.

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3 years ago

What two elements are used to make a laser pen light?

irina1246 [14]

Answer:

Helium and Neon

Explanation:

6 0

2 years ago

Andre is packing for a move to Kentucky. He needs to pack 20 small paperback books whose total mass is 5 kg and some pillows who

storchak [24]

The pillows because it is larger than paperback books, in this situation just think about which is larger, I pretty sure Andre is not going to hold the boxes in his hand.
hope I helped

6 0

3 years ago

Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13

Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

$\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\$

$=\frac{1571}{10}\\\\=157.1$

Calculating the standardn:

$\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\$

Please find the correct equation in the attached file.

$=28.195$

For point a:

$=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\$

For point b:

$=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M$

For point c:

$= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\ M$

It is calculated by using the slope value that is $1.75 \times 10^9 M^{-1}$ . The slope value $1.75 \times 10^9 M^{-1}$ is ambiguous.

7 0

2 years ago

How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of

alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

$E = k \frac{q}{r^{2} }$

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

$E = k \frac{q}{r^{2} }$

q = E x r²

q = 1250 N/C x 0.205m²

8.99 x 10⁹ Nm²/C²

q = 5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

$n = \frac{q}{e}$

n = 5.84 x 10⁻⁹ C

1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0

3 years ago