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3 years ago

What does nm measure in the spectra?

Chemistry

2 answers:

timama [110]3 years ago

6 0

Answer: Nanometer

Explanation:

The nanometre (international spelling as used by the International Bureau of Weights and Measures; SI symbol: nm) or nanometer (American spelling) is a unit of length in the metric system, equal to one billionth (short scale) of a metre (0.000000001 m).

...

Nanometer

imperial/US units 3.2808×10−9 ft 3.9370×10−8 in

coldgirl [10]3 years ago

3 0

Answer:

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The hcl(g) molecule has a bond length of 127 pm and a dipole moment of 1.08

Roman55 [17]

Let us assume that this molecule is 100 percent ionic. In that case, the charges are distinguished by a bond length.

h = Q × r

= (160 × 10⁻¹⁹ c) (127 × 10⁻¹² m) (10 / 3.336 × 10⁻³⁰ cm)

= 6.09 D.

The actual dipole moment is = 1.08 D

Therefore, the percent ionic character is,

= 1.08 D / 6.09 D × 100

= 17.7 %.

6 0

3 years ago

If an atom gains an electron it is known as what ? And has a negative charge

bekas [8.4K]

If an atom gains an electron and gets a negative charge because of it, it is a negatively charge ion AKA an anion.

6 0

3 years ago

What is the final temperature when 150.0 g of water at 90.0 °c is added to 100.0 g of water at 30.0 OC? Note that C, of water ca

levacccp [35]

Answer : The final temperature is, $337.8K$

Explanation :

$Q_{absorbed}=Q_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water at $90^oC$ = 150 g

$m_2$ = mass of water at $30^oC$ = 100 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of lead = $90^oC=273+90=363K$

$T_2$ = temperature of water = $30^oC=273+30=303K$

$c_1\text{ and }c_2$ = same (for water)

Now put all the given values in equation (1), we get

$150\times (T_{final}-363)=-[100\times (T_{final}-303)]$

$T_{final}=337.8K$

Therefore, the final temperature is, $337.8K$

8 0

4 years ago

Read 2 more answers

In order to separate the components of a binary mixture, the relative volatility should be a) less than unity b) equal to unity

Tresset [83]

Answer: Option (c) is the correct answer.

Explanation:

A binary mixture is defined as the mixture which contains two components in the aqueous medium. The two components are solute and solvent.

And, volatility is defined as the ability of a liquid solution or substance to readily change into vapors.

For a binary solution the expression for relative volatility is as follows.

$\alpha = \frac{\frac{y_{i}}{x_{i}}}{\frac{y_{j}}{x_{j}}}$ = $K_{i}/K_{j}$

where, $\alpha$ = relative volatility of more volatile component i

$y_{i}$ = vapor-liquid equilibrium concentration of component i in the vapor phase

$x_{i}$ = vapor-liquid equilibrium concentration of component i in the liquid phase

$y_{j}$ = vapor-liquid equilibrium concentration of component j in the vapor phase

$x_{j}$ = vapor-liquid equilibrium concentration of component j in the liquid phase

So, when $\alpha$ > 1 then separation by distillation is easier in nature.

Thus, we can conclude that in order to separate the components of a binary mixture, the relative volatility should be greater than unity.

8 0

3 years ago

The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C

Lana71 [14]

Explanation:

First, we will calculate the molar mass of $C_{6}H_{6}$ as follows.

Molar mass of $C_{6}H_{6}$ = $6 \times 12 + 6 \times 1$

= 78 g/mol

So, when 2 mol of $C_{2}H{6}$ burns, then heat produced = 6542 KJ

Hence, this means that 2 molecules of $C_{6}H{6}$ are equal to $78 \times 2 = 156 g$ of $C_{6}H_{6}$ burns, heat produced = 6542 KJ

Therefore, heat produced by burning 5.5 g of $C_{6}H{6}$ =

$6542 kJ \times \frac{5.5 g}{156 g}$

= 228.97 kJ

= 228970 J (as 1 kJ = 1000 J)

It if given that for water, m = 5691 g

And, we know that specific heat capacity of water is 4.186 $J/g^{o}C$ .

As, Q = $m \times C \times (T_{f} - T_{i})$

228970 J = $5691 g \times 4.184 J/g^{o}C \times (T_{f} - 21)
^{o}C$

$T_{f} - 21^{o}C = 9.616^{o}C$

$T_{f} = 30.6^{o}C$

Thus, we can conclude that the final temperature of the water is $30.6^{o}C$ .

4 0

3 years ago