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3 years ago

As you move from left to right across the periodic table, what happens (in terms of metals/nonmetals, etc.)?

Physics

2 answers:

Andru [333]3 years ago

8 0

Answer:

Hope it helps;)

Have a good day

stepan [7]3 years ago

8 0

As you move from left to right across the periodic table, you start to leave the metals and run into the metalloids like boron and silicon. As you keep going right, you eventually run into the non metals like carbon, nitrogen, and oxygen. Metals and non metals are separated by a diagonal line.

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What is the volume of a 1.2kg and displaced 1.0g/cm3

Zinaida [17]

Mass = 1.2 kg = 1200 grams.

Volume = mass/density = 1200 cm3.

Hope this helps!

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3 years ago

the scores of players on a golf team are shown in the table. the teams combined score was 0 what was travis's score?

Alona [7]

Answer:

what table?

Explanation:

3 0

3 years ago

Read 2 more answers

Lightning produces a maximum air temperature on the order of 104K, whereas a nuclear explosion produces a temperature on the ord

gtnhenbr [62]

Answer:

tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

$2.898\times 10^{-10}\ \text{m}$ x-ray region

Explanation:

T = Temperature

b = Constant of proportionality = $2.898\times 10^{-3}\ \text{m K}$

$\lambda$ = Wavelength

$T=10^4\ \text{K}$

From Wein's law we have

$\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}$

The wavelength of the radiation will be $2.898\times 10^{-7}\ \text{m}$ and it is in the ultraviolet region.

$T=10^7\ \text{K}$

$\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}$

The wavelength of the radiation will be $2.898\times 10^{-10}\ \text{m}$ and it is in the x-ray region.

5 0

3 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

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3 years ago

Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. Exp

VARVARA [1.3K]

The unbalanced reaction is

a NH₄NO₃ ⇒ b N₂ + c O₂ + d H₂O

where a, b, c, and d are unknown constants.

Count how many times each element appears on either side of the reaction.

• reactants:

N = 2a, H = 4a, O = 3a

• products

N = 2b, O = 2c + d, H = 2d

Now we solve the system of equations,

2a = 2b … … … [1]

4a = 2d … … … [2]

3a = 2c + d … … … [3]

From [1] we immediately have

a = b

In [2], we get

2a = d

and substituting for d in [3] gives

3a = 2c + 2a

a = 2c

Let c = 1; then

• a = 2×1 = 2

• b = 2

• d = 2×2 = 4

So, the balanced reaction is

2 NH₄NO₃ ⇒ 2 N₂ + O₂ + 4 H₂O

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2 years ago