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3 years ago

Why only electrons stick to oil drop in millikan's experiment however protons are also present there?

Chemistry

1 answer:

marishachu [46]3 years ago

5 0

Explanation:

probably because, X rays are used to ionise the gas molecules, which is loss of electrons, these electrons are absorbed by oil drops

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Two standard methods are used to express the relation between protein concentration and its absorbance at 280 nm. One of these m

TEA [102]

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2 years ago

This diagram shows the forces acting on a car. The center dot represents the car, and the arrows represent the forces acting on

Mariulka [41]

Answer : The correct option is B- Towards the left.

Solution : Given,

Force on car in upward direction, $F_1$ = 6000 N

Force on car in downward direction, $F_2$ = 6000 N

Force on car in right direction, $F_3$ = 5000 N

Force on car in left direction, $F_4$ = 6000 N

Force : Force is equal to the mass multiplied by the acceleration. The force is directly proportional to the acceleration.

Formula : F = m × a

where, F is force, m is mass and a is acceleration.

From the given diagram we conclude that the $F_1$ and $F_2$ are in opposite direction and the force acting on car are same. So, their net force is zero.

And in case of $F_3$ and $F_4$ , the net force will not be zero because force acting on car are different. The car will accelerate towards left direction because $F_4$ is higher than the $F_3$ .

So, the car will accelerate towards left direction.

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2 years ago

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

Katen [24]

Answer:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

Explanation:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

$Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}$

The pKa is

$-Log (Ka) = -Log (6.2x10^{-8}) = 7.2$

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.

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3 years ago

Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with

VladimirAG [237]

Answer:

7.3 g (NH₄)₃PO₄

Explanation:

The balanced equation for the reaction is:

H₃PO₄ + 3 NH₃ ----> (NH₄)₃PO₄

To find the mass of ammonium phosphate ((NH₄)₃PO₄) produced, you need to (1) convert grams NH₃ to moles NH₃ (via the molar mass from the periodic table), then (2) convert moles NH₃ to moles (NH₄)₃PO₄ (via mole-to-mole ratio from balanced equation), and then (3) convert moles (NH₄)₃PO₄ to grams (NH₄)₃PO₄ (via molar mass from periodic table). Make sure to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs because the given value (2.5 grams) has 2 sig figs.

Molar Mass (NH₃): 14.01 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.034 g/mol

Molar Mass ((NH₄)₃PO₄):

3(14.01 g/mol) + 12(1.008 g/mol) + 30.97 g/mol + 4(16.00 g/mol)

Molar Mass ((NH₄)₃PO₄): 149.096 g/mol

2.5 g NH₃ 1 mole NH₃ 1 mole (NH₄)₃PO₄ 149.096 g
--------------- x -------------------- x --------------------------- x --------------------------
17.034 g 3 moles NH₃ 1 mole (NH₄)₃PO₄

= 7.3 g (NH₄)₃PO₄

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1 year ago

What is the Kelvin temperature for 30°C? 130 K 30 K 273 K 293 K 303 K

pishuonlain [190]

The answer is 303 K, to find K you have to add 273 to the °C

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2 years ago