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3 years ago

Synthetic, or man-made, elements typically exist for only a short time

1 answer:

8 0

This is true. Elements past lead are radioactive, because the repulsive force of the protons cannot be overpowered by the “gluing” ability of neutrons (remember, likes repel). As more and more protons are added, generally, the elements become more unstable; for example, Bismuth, right next to lead on the Periodic Table, is radioactive, but the half life of this element is about a billion times longer than the current age of the universe, but Oganesson, element number 118, has a half life of fractions of a second.

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A 10.0 g sample of a hydrate was heated until all the water was driven off. The mass of the anhydrous product remaining was 8.00

chubhunter [2.5K]

Answer:

20%

Explanation:

10.0 = 100%

8.00 = 80%

100-80= 20

3 0

3 years ago

The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of ene

sesenic [268]

Answer:

$T_2=98.5^{\circ}$

Explanation:

Given that,

The specific heat of water is 4.184Jg°C

Mass, m = 600 g

Initial temperature, T₁ = 75.5°C

We need to find the final temperature. We know that heat absorbed is given by :

$Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}$

So, the final temperature is equal to $98.5^{\circ}$ .

3 0

3 years ago

Calculate the number of moles when 3.00l of a gas at 814 mmhg and 25 °c?

Mariulka [41]

Volume=V=3L
Pressure=P=814mmHg
Temp=T=25°C

No of moles=n

$\\ \rm\rightarrowtail PV=nRT$

$\\ \rm\rightarrowtail n=\dfrac{PV}{RT}$

$\\ \rm\rightarrowtail n=\dfrac{814(3)}{25R}$

$\\ \rm\rightarrowtail n=2442/25R$

$\\ \rm\rightarrowtail n=97.7Rmol$

5 0

2 years ago

9) After lab, all of Darrel’s friends looked at his data and laughed and laughed. They told him that he was 30.8% too low in the

zaharov [31]

If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.

In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.

So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC

3 0

4 years ago

Read 2 more answers

How many milliliters of a 1M nitric acid solution are required to prepare 60mL of 6.7M solution?

Free_Kalibri [48]

Answer:

the number of milliliters of a 1M is 402mL

Explanation:

The computation of the number of milliliters could be determined by using the following formula

As we know that

$V_1\times M_1 = V_2\times M_2$

where,

V_1 and V_2 are the starting and final volumes

And, the M_1 and M_2 are the starting and the final molarities

Now the V_1 is

$V_1 \times 1M = 60mL \times 6.7M$

So, the V_1 is 402mL

Hence, the number of milliliters of a 1M is 402mL

4 0

3 years ago