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zlopas [31]
3 years ago
5

Which diagram best represents light being reflected after striking the flat surface of a mirror

Chemistry
1 answer:
masya89 [10]3 years ago
5 0
It 1
cause yea it is im doing it in class and the answers is 1

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I NEED HELP ASAP!!<br> I DOMT KNOW HOW TO DO THIS
KatRina [158]
What part of it are you confused about
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3 years ago
What can you infer about copper's properties?
tiny-mole [99]

Answer:

good electrical conductor

lustre in appearance

7 0
3 years ago
How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602
hichkok12 [17]

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as 1.602 \times 10^{-19} C. Hence, we will calculate its number of moles as follows.

          Moles = \frac{mass}{\text{molar mass}}

                     = \frac{0.7 \times 1000 g}{7 g/mol}

                     = 100 mol

According to mole concept, there are 6.023 \times 10^{23} atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

        No. of atoms = 100 \times 6.023 \times 10^{23}

                               = 6.023 \times 10^{25} atoms

Since, it is given that charge on 1 atom is as follows.

                     3 \times 1.602 \times 10^{-19}C

                    = 4.806 \times 10^{-19}C

Therefore, charge present on 6.023 \times 10^{25} atoms will be calculated as follows.

    6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C

            28.95 \times 10^{6}C

Thus, we can conclude that a positive charge of 28.95 \times 10^{6}C is in 0.7 kg of lithium.

3 0
3 years ago
A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist
REY [17]

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

Explanation:

Determination of sign of q

Temperature of the water bath before the reaction = 25 °C

Temperature of the water bath after the completion of the reaction = 28 °C

After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

3 0
3 years ago
Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
3 years ago
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