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4 years ago

Conductive definition science term

Chemistry

1 answer:

Volgvan4 years ago

5 0

having the property or capability of conducting.

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Plz help .....................

Pavel [41]

Answer:

1. conduction

2. radiation

3.convection

4. conduction?

5. radiation

6. convection

7. convection

8. convection

9. radiation

10.radiation

(sorry if they aren't all correct)

6 0

3 years ago

Which of the following is the best definition of nuclear fission?

Reika [66]

Answer:

Explanation:

Nuclear ﬁssion is a process where the nucleus of an atom is split into two or more smaller nuclei, known as ﬁssion products.

8 0

3 years ago

Read 2 more answers

What is the difference between deuterium and tritium

disa [49]

Deuterium has one proton and one neutron, 12H, while tritium has one proton and two neutrons, 13H.

3 0

3 years ago

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

Neporo4naja [7]

Answer:

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Explanation:

Convert the given original molarity to molar as follows.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Consider the following serial dilutions.

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

Molarity of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is diluted to 250 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

Molarity of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM :

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

Molarity of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM :

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

Molarity of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is diluted to 500 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 10 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

Molarity of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

7 0

3 years ago

explain the difference between a neutral atom an isotope and an ion. use a specific element and explain the three possible forms

ikadub [295]

Atoms are made of three types of sub atomic particles, they are the electrons, neutrons and protons.
Electrons are negatively charged and they are orbiting around the nucleus in energy shells. Protons are positively charged , neutrons are neutral and have no charge, both neutrons and protons are located in the nucleus.
In atoms the atomic number is the number of protons the atom consists of and mass number is the sum of protons and neutrons in the nucleus.

In a neutral atom, the number of electrons and protons are equal, this means that the number of negative charges and positive charges are equal.
lets take Cl as an example. Its atomic number is 17.
number of protons and charge - +17
number of electrons with charge - -17
overall chage sum of the charges = + 17 -17 = 0
therefore no charge, hence its neutral.

isotopes are the same element having the same atomic number but different mass numbers.
³⁵Cl -protons - 17 electrons - 17 neutrons 18 neutrons
³⁷Cl - protons - 17 electrons - 17 neutrons 20 neutrons
Both are same element with same number of protons and electrons, however the number of neutrons are different.
Since protons + neutrons = mass of the atom
and when the number of neutrons change - mass of atoms too change
hence, ³⁵Cl has a mass of 35 and ³⁷Cl has a mass of 37. both of these are called isotopes

Ions are atoms which have either gained an electron or lost an electron and are charged.
metals such as K(atomic number - 19) have one valence electron in the outer shell, to gain the configuration of a complete octet, K needs to give out this one electron.
then number of electrons 19-1 = -18
number of protons = +19
overall charge = +1
K is one electron less therefore becomes positively charged and called a cation- K⁺
if we take Cl, atomic number - 17
Cl has 7 valence electrons, to gain a complete octet in the outer shell it needs to have 8 electrons, therefore it needs one more electron to complete the outer shell.
Therefore it gains one electron,
number of electrons 17+1 = -18
number of protons = 17
overall charge = -1
with an extra electron, Cl becomes negatively charged, a negative ion called anion - Cl⁻

3 0

3 years ago