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3 years ago

The photoelectric effect occurs when _____ are emitted from metal when the metal is struck by light of certain frequencies.

Chemistry

2 answers:

zavuch27 [327]3 years ago

8 0

Answer:

Electrons

Explanation:

Edge 2021

Ainat [17]3 years ago

4 0

Answer:

electrons

Explanation:

The photoelectric effect occurs when electrons are emitted from metal when the metal is struck by light of certain frequencies.

Some of the applications of this effect include photomultipliers (which are a key component in spectroscopy instruments) and night vision devices.

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Which unit would be used in energy calculations in thermodynamics?

prisoha [69]

Answer:

kJ/mol

Explanation:

4 0

3 years ago

How many moles of water are produced if 5.43 mol PbO2 are consumed? ANSWER: 10.9

Arada [10]

Answer: 10.9 mol.

Explanation:

To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.

Using cross multiplication:

1.0 mole of PbO₂ → 2.0 moles of H₂O

5.43 moles of PbO₂ → ??? moles of water

The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.

4 0

3 years ago

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Put these numbers from greatest to least precision 0.10 g/mL 0.0447g/mL 0.0700g/mL

horrorfan [7]

I am going to go with,

0.10 g/mL
0.0700 g/mL
0.0447 g/mL

I don't know if this is the correct answer, but I am 80% sure that it may be.
:) :)

6 0

3 years ago

Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

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4 years ago

The difference between skeletal and cardiac muscles is that the cardiac muscle is involuntary and cannot be actively controlled.

Virty [35]

True ( sorry if i’m wrong )

5 0

3 years ago

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