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2 years ago

Do you think this one experiment is enough to change the scientific theory regarding the speed of light?

1 answer:

7 0

No. The speed of light does change. You mean speed of light in a vacuum. Measuring that value is elementary, in a 2-way measurement, as you show.

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A system of gas at low density has an initial pressure of 1.90 × 10 5 1.90×105 Pa and occupies a volume of 0.18 m³. The slow add

Bad White [126]

Answer:

ΔU = -6.2 × 10⁴ J

Explanation:

The system absorbs 965 J of heat, that is, q = 965 J.

The work (w) can be calculated using the following expression.

w = -P . ΔV

where,

P is the external pressure

ΔV is the change in the volume

w = - (1.90 × 10⁵ N/m²) × (0.51 m³ - 0.18 m³) = -6.3 × 10⁴ J

The change in the internal energy (ΔU) is:

ΔU = q + w = 965 J + (-6.3 × 10⁴ J) = -6.2 × 10⁴ J

8 0

3 years ago

Help it's science so help

bija089 [108]

1.b
2.b
3.a
3 was the the opposite of the scenario stated

7 0

3 years ago

GIVING BRAINLY, 5 STARS, FOLLOW AND HEARTS<br>as majority of substances heat up, they:

melamori03 [73]

Answer:

Expand

Explanation:

When you heat up a substance it expands

3 0

2 years ago

What is the mass of 2.40 moles of magnesium chloride, mgcl2

LiRa [457]

Hey there!

MgCl₂

Find molar mass of magnesium chloride.

Mg: 1 x 24.305

Cl: 2 x 35.453

--------------------

95.211 grams

One mole of magnesium chloride has a mass of 95.211 grams.

We have 2.40 moles.

2.40 x 95.211 = 228.5

To 3 sig figs this is 229.

The mass of 2.40 moles of magnesium chloride is 229 grams.

Hope this helps!

4 0

3 years ago

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c

uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

$I_{2} (g) + Cl_{2} (g) -----> 2 ICl (g)$

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

$K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}$

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = $\frac{(2x)^{2}}{(0.437-x) (0.269-x)}$

81.9 = $\frac{(2x)^{2}}{x^{2} -0.706x + 0.118}$

$(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]$

$4x^{2} = 81.9x^{2} -57.8x +9.66$

$77.9x^{2} -57.8x+9.66 = 0$

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

$[ICl]_{eq} = 2 ( 0.254) = 0.508 M$

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0

3 years ago