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2 years ago

How do different types of waves make particles of matter move

1 answer:

6 0

That depends on the wave, if you're talking about sound, it makes matter move in a similar wavelength as them, a mountainous shape. Light however would make whatever matter it hits start to move in the same direction as the light's angle of approach.

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3) The spectrum of the unknown mixture is a mixture of two substances among the five. What are these two substances

Neporo4naja [7]

Answer:

a. H + d. Na

Explanation:

This is Mixture of H and Na. Because of H and in the emission and Absorption spectrum of H Total 5 lines [Red, Green, Blue and Purple (2)) are present and in the spectrum of Na Two yellow lines are present. In the mixture all this lines are are also 5 + Present. Total lines are 5+2= 7 (H+ Na)

Therefore, the answer is a. H + d. Na

3 0

3 years ago

What is pH?

sergij07 [2.7K]

PH = -log[H+]
That is, pH is the negative logarithm of the hydrogen ion concentration.

pOH = -log[OH-]
pOH is the negative logarithm of the hydroxide ion concentration.

The correct answer to your question is a. The negative logarithm of the hydrogen ion concentration.

5 0

2 years ago

P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles, R is the gas constant (0.0821 L

yKpoI14uk [10]

Answer:

The mass of the neon gas m = 1.214 kg

Explanation:

Pressure = 3 atm = 304 k pa

Volume = 0.57 L = 0.00057 $m^{3}$

Temperature = 75 °c = 348 K

Universal gas constant = 0.0821 $\frac{L . atm}{mol K}$

We have to change the unit of this constant. it may be written as

Universal gas constant = 8.314 $\frac{KJ}{mol K}$

Gas constant for neon = $\frac{8.314}{20}$ = 0.41 $\frac{KJ}{kg K}$

From ideal gas equation,

P V = m R T ------- (1)

We have all the variables except m. so we have to solve this equation for mass (m).

⇒ 304 × $10^{3}$ × 0.00057 = m × 0.41 × 348

⇒ 173.28 = 142.68 × m

⇒ m = 1.214 kg

This is the mass of the neon gas.

4 0

2 years ago

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

What are kind and number of atoms in ring portion in the haworth structure of glucose

monitta

Answer:

The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.

3 0

2 years ago

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