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1 year ago

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X-rays have a wavelength small enough to image individual atoms, but are challenging to detect because of their typical frequenc

y. Suppose an X-ray camera uses X-rays with a wavelength of 8.38nm . Calculate the frequency of the X-rays. Round your answer to 3 significant digits.

1 answer:

Cerrena [4.2K]1 year ago

7 0

Explanation:

Given: $\lambda = 8.38\:\text{nm} = 8.38×10^{-9}\:\text{m}$

Its frequency $\nu$ is defined as

$\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{8.38×10^{-9}\:\text{m}}$

$\:\:\:\:= 3.58×10^{16}\:\text{Hz}$

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Unit Conversion Help Thank you

AlekseyPX

Answer : 1721.72 g/qt are in 18.2 g/cL

Explanation :

As we are given: 18.2 g/cL

Now we have to convert 18.2 g/cL to g/qt.

Conversions used are:

(1) 1 L = 100 cL

(2) 1 L = 1000 mL

(3) 1 qt = 946 qt

The conversion expression will be:

$\frac{18.2g}{1cL}\times \frac{100cL}{1L}\times \frac{1L}{1000mL}\times \frac{946mL}{1qt}$

$=1721.72\text{ g/qt}$

Therefore, 1721.72 g/qt are in 18.2 g/cL

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2 years ago

Perform the following

Ghella [55]

Answer:

1.85 × 10⁻⁶

Explanation:

0.0003 ÷ 162 = 1.851851852 × 10⁻⁶ ⇒ 1.85 × 10⁻⁶

Hope that helps.

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2 years ago

A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.

lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

$\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol$

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

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2 years ago

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

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2 years ago

A sample of n2 gas occupies 2.40 l at 20°c. if the gas is in a container that can contract or expand at constant pressure, at wh

Bas_tet [7]

From Charle's law the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant pressure.
Therefore';
V1/T1=V2/T2
Where; V1 = 2.40 l, T1 = 273 +20= 293 K, V2 = 4.80, and T2= ?
2.4/293= 4.8/T2
T2= (4.8×293)/2.4
= 586 K or 313° C

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2 years ago

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