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SIZIF [17.4K]
3 years ago
8

Which most accurately describes the universe as we currently describe it?

Chemistry
1 answer:
weeeeeb [17]3 years ago
4 0

Answer: The universe consists of all physical objects on earth in space

Explanation:

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Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficien
Vadim26 [7]

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

7 0
3 years ago
A 232-lb fullback runs the 40-yd dash at a speed of 19.8 ± 0.1 mi/h.
Neporo4naja [7]

Answer:

(a)  7.11 x 10⁻³⁷ m

(b)  1.11 x 10⁻³⁵ m

Explanation:

(a)  The de Broglie wavelength is given by the expression:

λ = h/p = h/mv

where h is plancks constant, p is momentum which is equal to mass times velocity.

We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.

v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s

m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg

λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m

(b) For this part we have to use the uncertainty principle associated with wave-matter:

ΔpΔx > = h/4π

mΔvΔx > = h/4π

Δx = h/ (4π m Δv )

Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.

Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )  

     = 0.045 m/s

Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )

     = 1.11 x 10⁻³⁵ m

This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.

5 0
3 years ago
Hey need help. this is ONE question a 4 part one though. pls help
shusha [124]
I don’t get it. What is the question asking?
3 0
3 years ago
Read 2 more answers
Why can gasses change volume?​
soldi70 [24.7K]

Answer:

b. I've seen a question like this before lol

7 0
3 years ago
Read 2 more answers
If the rate of a reaction increases by a factor of 64 when the concentration of reactant increases by a factor of 4, what is the
igomit [66]

The reaction is of order three with respect to the reactant.

<h3>Explanation</h3>

The rate of a reaction of order n about a certain reactant is proportion to the concentration of that reactant raised to the n-th power. This is true only if  concentrations of any other reactants stay constant in the whole process.

In other words, Rate = constant × [Reactant]ⁿ, Rate ∝ [Reactant]ⁿ. (The symbol "∝" reads "proportional to".)

In this question,

[4 × Reactant]ⁿ ÷ [Reactant]ⁿ = 64.

In other words, 4ⁿ = 64, where n is the order of the reaction with respect to this reactant.

It might take some guesswork to find the value of n. Alternatively, n can be solved directly with a calculator using logarithms. Taking natural log of both sides:

\ln{4^n} = \ln{64}\\n\; \ln{4} = \ln{64}\\n = \frac{\ln{64}}{\ln{4}} = 3.

Evaluating \ln(64) / \ln(4) on Google or on a calculator with support for ln (the natural log) will give the value of n- no guesswork required.

n = 3. Therefore, the reaction is of order three with respect to this reactant.

3 0
3 years ago
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