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3 years ago

The isotopes of an element all have the same (atomic, mass) number, but they have different (atomic,mass)numbers.

1 answer:

4 0

<h2>The isotopes of an element all have the same __(atomic, mass) __number, but they have different __(atomic,mass)__numbers.</h2>

Explanation:

The isotopes of an element all have the same __atomic number __, but they have different __mass __numbers.

The isotopes have same atomic number that is :

Same number of electrons
Same number of protons
same electronic configuration
same valence electrons
same valency
same symbol

The isotopes have different mass number that is :

They differ in number of neutrons .

For example : Isotopes of hydrogen are : H₁¹ , H₁² , H₁³

isotopes of Oxygen is : O¹⁶ , O¹⁷, O¹⁸

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A bottle of soda has a PH of 3.

BaLLatris [955]

Answer:

The new pH of the solution $= 6$

Explanation:

Given

pH of bottle of Soda is $3$

As we know that

pH $= -$ log [H+]

We will derive the hydrogen ion concentration when pH is $3$

$3 = -$ log [H+]

Taking anti-log on both the sides, we get -

H+ ion is equal to $10^{-3}$

When the hydronuim ion concentration reduces by $\frac{1}{1000}$ of the original concentration, the new hydrogen ion concentration becomes

$\frac{1}{1000000} = 10^{-6}$

The new pH is

$= - log [ 10^{-6}]\\= 6$

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3 years ago

PCL was mixed with gelatin to make a blend for elctrospun fibrous scaffold encapsulating growth factor that was admixed in the p

shusha [124]

Answer:

Explanation:

First PCL to form a scaffold, combined with gelatin.

They are made by first three forms A) made by just PCL.B) made by gelatin ratio PCL is 3:1 and last is C) made by gelatin PCL.

The decoration rate is 1%, 25% and 50% respectively.

<em>A) Growth factor is stuck in 21 days, and can not spread. In this case in vivo experiment for 7 days used highly degradable scaffold use the ability to break down due to decomposition in vivo degradation rate depends on the scaffold's acid byproduct impact.</em>

B) the amount of scaffolded degradation.

First, with scaffold A.

10 mg scaffold weight A= 1 per cent degradation.

Following degradation wt is 9.967=? Degradation per cent.

So, degradation (9.967* 1)/10= 0.9967 per cent.

Likewise for C) scaffold c 10 mg wt. Loss of 50 per cent.

After wt 8.33=? Degradation per cent.

Degradation (8.33* 50)/10=41.65 per cent.

Scaffold c degrades significantly, since the loss of wt is even greater.

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4 years ago

What four elements did the ancient greeks think the world was made of ?

Nezavi [6.7K]

The ancient greeks thought that the earth was composed of earth, water, fire, and air.

Hope this helps you!

5 0

3 years ago

Read 2 more answers

What did the experiments of scientists after John Dalton reveal about his

EastWind [94]

A. Dalton's theory that atoms could not be divided was incorrect

3 0

3 years ago

Calculate the volume of carbon dioxide at 20.0°C and 0.941 atm produced from the complete combustion of 4.00 kg of methane. Comp

tankabanditka [31]

Answer:

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

Explanation:

Methane

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Mass of methane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of methane = $\frac{4000 g}{16 g/mol}=250 mol$

According to reaction, 1 mole of methane gives 1 mole of carbon dioxide gas,then 250 moles of methane will give :

$\frac{1}{1}\times 250 mol=250 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 250 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{250 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6390.89 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

Propane

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Mass of propane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of propane = $\frac{4000 g}{44 g/mol}=90.91 mol$

According to reaction, 1 mole of propane gives 3 mole of carbon dioxide gas,then 90.91 moles of propane will give :

$\frac{3}{1}\times 90.91 mol=272.73 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 272.73 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{272.73 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6,971.95 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6,971.95 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of methane = 6390.89 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of propane = 6,971.95 Liters.

6390.89 Liters < 6,971.95 Liters

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

7 0

3 years ago