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Y_Kistochka [10]
3 years ago
14

The term Reuther Ford gave to the positively charged particles in the nucleus of an antom was

Chemistry
1 answer:
lubasha [3.4K]3 years ago
4 0
The term Rutherford gave to the positively charged particles in the nucleus of an atom was/is Proton. 

Hope this helps!
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Keiko needs 100mL of a 5% acid solution for a science experiment. She has available a 1% solution and a 6% solution. How many mi
4vir4ik [10]

Answer:

Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution

Explanation:

There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:

1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%

2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%

When clearing X and Y :

(X mL 6%    +   Y mL 1%      =    100 mL 5%) (-15,7)

X mL 94%  +   Y mL 99%   =   100 mL 95%

_______________________________

 -                      Y  0.83       =   16.5

                        Y                =   19.9 mL 1% solution

Replace Y in anyone equation and X = 80 mL 6% solution

I hope to see been helpful

7 0
3 years ago
Read 2 more answers
The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
How many moles of He are in 16 g of the element? 1.0 mol 4.0 mol 8.0 mol 16 mol
Lady bird [3.3K]

Answer: 4 Mole

Explanation:

16/4 = 4

4 0
3 years ago
explain how both hydrogen and carbon have achieved stability by bonding with each other to form methane ​
Naya [18.7K]

A carbon iota can bond with four other iotas and is just like the four-hole wheel, whereas an oxygen iota, which can bond only to two, is just like the two-hole wheel.

4 0
3 years ago
Read 2 more answers
What do all organic substances have in common
motikmotik
All organic substances contain carbon atoms.
6 0
3 years ago
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