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lys-0071 [83]
3 years ago
10

Which of the following is NOT a "big idea" in chemistry?

Chemistry
1 answer:
aalyn [17]3 years ago
8 0
B but im not sure if im 100% correcr
You might be interested in
How many grams of octane (C8H18) must be burned to produce 300.0g of CO2?
nika2105 [10]

Answer:

m_{C_8H_{18}}=85.67gC_8H_{18}

Explanation:

Hello there!

In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:

m_{C_8H_{18}}=300.0gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molC_8H_{18}}{25molO_2}*\frac{114.22gC_8H_{18}}{1molC_8H_{18}}   \\\\m_{C_8H_{18}}=85.67gC_8H_{18}

Regards!

6 0
3 years ago
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

6 0
3 years ago
What is the difference between osmosis and diffusion?
Scorpion4ik [409]
<span>B. Osmosis is movement of proteins, and diffusion is movement of water.
</span>
3 0
3 years ago
Read 2 more answers
Part A
Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

  • <u>7.33 × 10 ⁻³ c</u>

Explanation:

<u>1) Data:</u>

a) m = 9.11 × 10⁻³¹ kg

b) λ =  3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

<u>2) Formula:</u>

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

Solve for s:

  • s = h / (m.λ)

Substitute:

  • s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg ×  3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c =  3.00 10⁸ m/s

  • s =  2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

  • 2.06 × 10 ⁻³⁴ m.

Explanation:

<u>1) Data:</u>

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ =  ?

<u>2) Formula:</u>

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

  • λ = h / (m.s)

Substitute:

  • λ =  6.626×10⁻³⁴J.s / ( 0.0459 kg ×  70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.  

5 0
3 years ago
What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

7 0
3 years ago
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