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Alenkasestr [34]
2 years ago
7

True or false during reactions or light is needed carbon dioxide is converted to glucose

Chemistry
1 answer:
Helen [10]2 years ago
7 0

Answer:

True

Explanation:

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Which of the following statements is not true for an exothermic reaction? Question options: The products have a higher heat cont
dsp73

Answer:

The products have a higher heat content than the reactants.

Explanation:

The statement above is not true for an exothermic reaction because in an exothermic reaction heat is released to the surroundings. This simply means that the total energy of the products is less than that of the reactants.

5 0
3 years ago
A student pushes a 40-N block across the floor for a distance of 10 m.
Ganezh [65]
C I think but not due e
5 0
3 years ago
Summarize the characteristics of each type of element below
aniked [119]
What are the elements?
7 0
2 years ago
Drag each tile to the correct location on the image.
nikdorinn [45]

Answer:

[He]: 2s² 2p⁵.

[Ne]: 3s².

[Ar]: 4s² 3d¹⁰ 4p².

[Kr]: 5s² 4d¹⁰ 5p⁵.

[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².

Explanation:

  • Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.

  • [He]:

He contains 2 electrons fill 1s (1s²).

So, [He] can be written before the electronic configuration of 2s² 2p⁵.

  • [Ne]:

Ne contains 10 electrons fill (1s² 2s² 2p⁶).

So, [Ne] can be written before the electronic configuration of 3s².

  • [Ar]:

Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).

So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².

  • [Kr]:

Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).

So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.

  • [Xe]:

Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).

So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².

3 0
3 years ago
Read 2 more answers
. How many milliliters of 0.20 M HCl are needed to exactly neutralize 40. milliliters of 0.40 M KOH
Anuta_ua [19.1K]

Answer:

V_{HCl}=80mL

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

KOH+HCl\rightarrow KCl+H_2O

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

n_{HCl}=n_{KOH}\\\\M_{HCl}V_{HCl}=M_{KOH}V_{KOH}

Hence, we compute the volume of HCl as shown below:

V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL

Best regards.

5 0
3 years ago
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