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4 years ago

Which element has the HIGHEST ionization energy? (Hint: it's hardest to remove an electron from this element)

Chemistry

2 answers:

Bumek [7]4 years ago

5 0

Answer:

Argon

Explanation:

Tanya [424]4 years ago

5 0

C. Argon, trust me I know

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What was the control in Pasteur's experiment?

Sholpan [36]

Answer:

a straight neck flask to allow air to get in

Explanation:

I took the test eWe

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4 years ago

3. Which muscle tissue lacks striations?

Diano4ka-milaya [45]

Answer:

your answer is b and hope ur help and mark me brainlist

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3 years ago

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle

Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of $4.71\times 10^{-15}J$ . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

= De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of beta particle = $9.1094\times 10^{-31} kg$

$E_k$ = kinetic energy of the particle = $4.71\times 10^{-15}J$

Putting values in above equation, we get:

$\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}$

$\lambda = 6.762\times 10^{-12} m$

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

3 0

3 years ago

The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera

marshall27 [118]

Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius

8 0

3 years ago

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

4 0

2 years ago