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3 years ago

What is the empirical formula of a compound that is 40% sulfur and 60% oxygen by weight?

Chemistry

1 answer:

Zigmanuir [339]3 years ago

3 0

The formula would be so3

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During photosynthesis, the following reaction takes place: carbon dioxide + water + light energy → sugar + oxygen During cellula

Pachacha [2.7K]

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It cost Evan $17.70 to send 177 text messages. How many text messages did he send if he spent $19.10?

Explanation:

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2 years ago

(06.01 MC)

skelet666 [1.2K]

Answer:

Part 1. When the balloon is filled half of the way, and placed into the freezer, it will shrink. This happens because kinetic molecular theory tells us that a decrease in temperature decreases the kinetic energy of the gas molecules in the balloon. Viscous gases like hydrogen are less likely to shrink.

Part 2. When the balloon is placed out in the hot sun, most likely the balloon will swell and grow. This happens because the kinetic energy of the gas molecules increases due to solar radiation transforming into heat energy and then transforming into kinetic energy. Sticky gases like neon are more likely to grow.

Explanation:

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3 years ago

Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist

strojnjashka [21]

Answer:

Force of attraction = 35.96 $\times 10^{27}$ N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = $10^{-9}$ m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

$F = \dfrac{k \times q_1 q_2}{ \r^2}$ ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = $8.99 \times 10^9 \ Nm^{2}C^{-2}$

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

$F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }$

Force of attraction = 35.96 $\times 10^{27}$ N

6 0

3 years ago

Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients

Elena L [17]

Answer : The balanced chemical equation in a acidic solution is,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

First balance the main element in the reaction.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

Now balance oxygen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Now balance the charge.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}+2e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : $3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

7 0

3 years ago

One sphere has a radions of 181 cm, another has a radius of 5.01 cm. What is the difference in volume (in cubic centimeters) bet

PSYCHO15rus [73]

Answer:

$2.4*10^{7}cm^{3}$

Explanation:

First you should calculate the volume of a big sphere,so:

$V_{big}=\frac{4}{3}\pi r^{3}$

$V_{big}=\frac{4}{3}\pi (181cm)^{3}$

$V_{big}=2.4*10^{7}cm^{3}$

Then you calculate the volume of a small spehre, so:

$V_{small}=\frac{4}{3}\pi r^{3}$

$V_{small}=\frac{4}{3}\pi (5.01cm)^{3}$

$V_{small}=5.3*10^{2}cm^{3}$

Finally you subtract the two quantities:

$V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}$

$V_{big}-V_{small}=2.4*10^{7}cm^{3}$

5 0

3 years ago