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Ann [662]
2 years ago
5

H2O

Chemistry
2 answers:
Anna71 [15]2 years ago
8 0

Answer:

the answer is  A; atom

Explanation:

hope this helps you

erma4kov [3.2K]2 years ago
3 0
B) Compound, H and O are elements, when two elements bond together they become a compound
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C4H10 + 02 ➡️ H20 + CO2<br> balance the equation
Alexeev081 [22]

(Can I have Brainlist)?

Answer:

2C4H10 + 13O2 —> 8CO2 + 10H2O. Oxidation reaction

8 (4 moles CO2 per mole butane)

Explanation:

could be written C4H10 + 6 1/2 O2 —> 4CO2 + 5H2O

4 0
2 years ago
Heart, 5 stars, Brainiest, and 40 coins if right! Answer needed ASAP
kicyunya [14]

Answer:

His kinetic energy is converted into potential energy.

Explanation:

6 0
2 years ago
If the visible light spectrum is from 400 to 700 nm, would light with an energy of 2.79 x 10^-19 J be visible with the naked eye
sladkih [1.3K]

Answer:

713 nm. It is not visible with the naked eye.

Explanation:

Step 1: Given data

  • Energy of light (E): 2.79 × 10⁻¹⁹ J
  • Planck's constant (h): 6.63 × 10⁻³⁴ J.s
  • Speed of light (c): 3.00 × 10⁸ m/s
  • Wavelength (λ): ?

Step 2: Calculate the wavelength of the light

We will use the Planck-Einstein equation.

E = h × c / λ

λ = h × c / E

λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 2.79 × 10⁻¹⁹ J

λ = 7.13 × 10⁻⁷ m

Step 3: Convert "λ" to nm

We will use the relationship 1 m = 10⁹ nm.

7.13 × 10⁻⁷ m × (10⁹ nm/1 m) = 713 nm

This light is not in the 400-700 nm interval so it is not visible with the naked eye.

5 0
3 years ago
What is the correct name for OF2?
Gre4nikov [31]
OF2 is Oxygen difluoride
7 0
3 years ago
Read 2 more answers
A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5
matrenka [14]

Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

<em>1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.</em>

<em />

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

<em>0.263M of CH₃COOH is the concentration of the solution</em>

3 0
3 years ago
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