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Varvara68 [4.7K]

3 years ago

7

2. What is an example of a molecule with different types of atoms?

1 answer:

agasfer [191]3 years ago

4 0

Answer:

a compound

Explanation:

A molecule with multiple atoms is compound like water it's made up of 1 oxygen and two hydrogen.

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For case 1, what happens when an electron jumps from energy level 1 to energy level 3 in an atom?

BaLLatris [955]

Answer:

Case 1 (energy level): In an atom, an electron jumps from energy level 1 to energy level 3. ... The energy will increase.

Explanation:

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3 years ago

Calculate the energy of a photon of radiation with a frequency of 9.3x10^14Hz

Yuki888 [10]

Answer:

10.5!!!!!!!!!!!!!!!!!!!!!!

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3 years ago

A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden

Luda [366]

Answer:

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

Concentration of $MnO_4^-$ = 0.03721 M

Explanation:

A)

The reduction for $Cr_2O_7^{2-}$ is;

$Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

6 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $Cr_2O_7^{2-}$ reacted = $\frac{0.00747}{6}$

number of moles of $Cr_2O_7^{2-}$ reacted = 0.001245 mole

Concentration of $Cr_2O_7^{2-}$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $Cr_2O_7^{2-}$ = $\frac{0.001245}{40.15*10^{-3}}$

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

B)

The reduction for $MnO_4^-$ is;

$MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

5 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $MnO_4^-$ reacted = $\frac{0.00747}{5}$

number of moles of $MnO_4^-$ reacted = 0.001494 mole

Concentration of $MnO_4^-$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $MnO_4^-$ = $\frac{0.001494 }{40.15*10^{-3}}$

Concentration of $MnO_4^-$ = 0.03721 M

3 0

4 years ago

12500 J/ (106 g) (4C) reduce to one

Sloan [31]

The given expression is $\frac{12500 J}{(106g)(4^{0}C )}$

This expression denotes the specific heat of a substance which indicates a heat energy of 12500 J is involved in raising the temperature of 106 g of the substance by $4^{0}C$ . Generally, the units of specific heat are $\frac{J}{g.^{0}C }$

$\frac{12500 J}{(106g)(4^{0}C )}$ = $\frac{12500 J}{(106g)(4^{0}C } = 29.5 \frac{J}{g^{0}C }$

Therefore, $\frac{12500 J}{(106g)(4^{0}C )}$ when reduced to one unit will be 29.5 $\frac{J}{g^{0}C }$

5 0

3 years ago

Read 2 more answers

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lozanna [386]

B. An early form of chemistry that people used to try to turn metal.

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3 years ago