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guajiro [1.7K]
3 years ago
12

Why do scientists consider any hypothesis valuable?

Chemistry
1 answer:
Serjik [45]3 years ago
3 0

Answer:

I hope this answer is correct

Explanation:

A hypothesis requires no further investigation if it is proved by the experiment.

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Hockey is a sport in which players use hockey sticks to push a puck across smooth ice. At practice one day, three different-colo
motikmotik

Answer:

The purple hockey puck; it takes a stronger force to slow down a more massive object than to speed it up.

Explanation:

Hope this helps! :) Plz mark as brainliest if u can!

6 0
2 years ago
Which of the following lists the stages of mitosis in order
AnnyKZ [126]

Anaphase

Metaphase

Prophase

Telophase

Interphase

Prometaphase

Explanation:

4 0
3 years ago
Read 2 more answers
How many moles of a gas would occupy 22.4 Liters at 273 K and 1 atm?
Molodets [167]

Answer:

1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm

Explanation:

An ideal gas is a set of atoms or molecules that move freely without interactions. The pressure exerted by the gas is due to the collisions of the molecules with the walls of the container. The ideal gas behavior is at low pressures, that is, at the limit of zero density. At high pressures the molecules interact and intermolecular forces cause the gas to deviate from ideality.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 1 atm
  • V= 22.4 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T=273 K

Reemplacing:

1 atm* 22.4 L= n* 0.082 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1 atm* 22.4 L}{0.082 \frac{atm*L}{mol*K} *273 K}

n= 1 mol

Another way to get the same result is by taking the STP conditions into account.

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C (or 273 K) are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<u><em>1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm</em></u>

4 0
2 years ago
Which of the following statements about the ordinary IR spectroscopic regions are TRUE? 1. In general, the IR FUNDAMENTAL region
yaroslaw [1]

Answer: the statements in 1 and 2 are true of IR spectroscopic region.

1. In general, the IR FUNDAMENTAL region has a longer wavelength region than the region we call the ultraviolet (uv) region.

2. We can sense some of the frequencies of the FUNDAMENTAL region of the IR as heat

Explanation:

IR has energy value between 10^-5eV - 10^-2eVwhile

UV has energy value of 4eV - 300eV

IR has low photon energy and cannot alter atoms and molecules while UV has sufficient energy to iodize atoms therefore UV has a higher energy band.

Infrared light falls just outside the visible spectrum, beyond the edge of what we can see as red.

5 0
3 years ago
Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
3 years ago
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