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Reptile [31]
3 years ago
6

The temperature in the stratosphere gets warmer with altitude. True False

Chemistry
1 answer:
larisa [96]3 years ago
7 0

Answer: False

Explanation: the temperature increases with altitude

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Which of the following type of radiation is represented by 0 over negative 1 e? A.alpha emission B.beta emission C.positron emis
Zolol [24]
The answer is B. beta emission. Alpha emission is a particle with 2 protons and 2 neutrons. Positron emission is represented by 0 over positive 1 e. Gamma rays is 0 over 0 γ.
7 0
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
Tris is a molecule that can be used to prepare buffers for biochemical experiments. It exists in two forms: Tris (a base) and Tr
ale4655 [162]

Solution :

For the reaction :

$\text{TrisH}^+ + H_2O \rightarrow \text{Trish}^- + H_3O^+$

we have

$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$

   $=\frac{x^2}{0.02 -x}$

  $=8.32 \times 10^{-9}$

Clearing $x$, we have $x = 1.29 \times 10^{-5} \text{ moles of acid}$

So to reach $\text{pH} = 7.8 (\text{pOH}= 14-7.8=6.2)$, one must have the $\text{OH}^-$ concentration of the :

$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$

So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.

$\text {n NaOH}=1.29 \times 10^{-5}+6.31 \times 10^{-7}$

            $= 1.35 \times 10^{-5} \text{ moles}$

Volume NaOH $= 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \text{ mL}}{1 \text{ mol}} = 0.0135 \text{ mL}$

Tris mass $H^+ = 0.02 \text{ mol} \times 157.6 \text{ g/mol}=3.152 \text{ g}$

Now to prepare the said solution we must mix:

$3.152 \text{ g Tris H} + 0.0135 \text{ mL NaOH} \ 1 M$ gauge to 1000 mL with water.

3 0
3 years ago
Which of the following is a reduction half-reaction?
Marta_Voda [28]

Solution : An oxidation reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between tow species an oxidaion reductin reaction is any chemical reaction in which the oxidation number of a molecule atom or ion changes by gaining or losing an electron

3 0
3 years ago
What is the concentration in milligrams per milliter of a solution containing 23.5 meq sodium chlorise per milliliter? mw nacl =
ValentinkaMS [17]

1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.

It is calculated in mg/ml.

The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.

Formula for calculating concentration in mg/ml is

Conc. (mg/ml) = M(eq) /ml ×  Molecular weight / Valency

Given

M(eq) NaCl/ ml = 23.5

Molecular weight pf NaCl = 58.5 g/mol

Valency = 1

Putting the values into the formula

Conc. (mg/ml) = 23.5 ×58.5/1

                       = 1374.75 mg/ml

Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Learn more about Concentration here brainly.com/question/14500335

#SPJ4

7 0
2 years ago
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