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3 years ago

Pls HELP AND EXPLAIN Can Jupiter support LIQUID water?

Chemistry

1 answer:

AURORKA [14]3 years ago

3 0

Answer:

The answer is yes, there is a small amount of water, but it is not ”on” Jupiter. It is in the form of water vapor in the cloud tops.

Explanation:

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__________ have an electric charge of -1.60 × 10-19 C (-e).

nignag [31]

Answer:

Electrons have an electric charge of -1.60 × 10-19 C

Explanation:

Neutrinos and Neutrons have 0 e charge.

Protons have an electric charge of 1.60 × 10-19 C

Historically, electrons were assigned a negative charge

6 0

3 years ago

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$

From the reaction, we conclude that

As, 1 mole of $KI$ react to give 1 mole of $AgI$

So, 0.0178 moles of $KI$ react to give 0.0178 moles of $AgI$

Thus,

Moles of AgI = Moles of $I^-$ anion = Moles of $Ag^+$ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

$\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}$

$\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M$

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0

3 years ago

The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime

krok68 [10]

Answer: The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = $2.83\times 10^{-3}$

a - x = amount left after decay process = $3.06\times 10^{-4}$

$559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=3.96\times 10^{-3}min^{-1}$

The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

6 0

3 years ago

How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of

alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

$E = k \frac{q}{r^{2} }$

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

$E = k \frac{q}{r^{2} }$

q = E x r²

q = 1250 N/C x 0.205m²

8.99 x 10⁹ Nm²/C²

q = 5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

$n = \frac{q}{e}$

n = 5.84 x 10⁻⁹ C

1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0

3 years ago

Alpha particles are helium nuclei. electromagnetic waves. electrons. neutrons.

Firdavs [7]

Alpha particles are helium nucleus ! so answer is A !

it is made by ionization of helium, when both the electrons are out , then two unit of positive charge is one helium and thus called alpha particle !

4 0

3 years ago