All categories

3 years ago

Ter the symbol for the isotope used in blood flow analyses.

Chemistry

1 answer:

blondinia [14]3 years ago

8 0

Answer:

The symbol of isotopes used for blood flow analysis is $\begin{array}{c}{141} \\ {58}\end{array} C e.$

<u>Explanation: </u>

Isotopes are the substances that exhibit the same atomic number but has a different mass number of an element.
The atomic number explains the number of protons present in the element and mass number explains the number of neutrons available in the element.
For blood flow analysis, the isotope element is cerium-141 and it is used in the chemical examination of blood flow particles.
Symbol used for this isotope is $\begin{array}{c}{141} \\ {58}\end{array} C e$ , where 141 indicates the amount of mass present and 58 indicates the proton number and 83 indicates neutron number present in that element.
The amount of mass in an atom is calculated by the sum of protons and neutrons present in it. Thus mass of isotope is 141 obtained by the sum of 58 protons and 83 neutrons present in that isotope.

You might be interested in

Which statement correctly compares an atom of an alkali metal with an atom of the alkaline earth metal next to it on the periodi

wolverine [178]

The atom of alkaline earth metal has one more proton and one more electron than the alkali metal that is next to it to the left on the periodic table. Then the atomic number of the alkali metal is one unit less than the atomic number of the alkaline earth metal next to it.

4 0

3 years ago

False answer it's F**** wrong

Aliun [14]

You know you can add comments right!!! <span />

8 0

3 years ago

A logical conclusion based on observations is called an inference.

Roman55 [17]

ANSWER: True
EXPLANATION: An inference is a logical conclusion based on observations.

Hope it helps u!

5 0

3 years ago

Read 2 more answers

A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance

sasho [114]

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$ , where K is the Coulomb´s constant ( $9x10^{9} N\frac{m^{2} }{C^{2} }$ ), $q_{1}$ = Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$ .

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$

Given that ri= 1.1nm= $1.1x10^{-9}m$ and rf= infinite distance

$W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J$

6 0

3 years ago

At 311 K, this reaction has a K c value of 0.0111 . X ( g ) + 2 Y ( g ) − ⇀ ↽ − 2 Z ( g ) Calculate K p at 311 K. Note that the

Aleks [24]

Answer:

$K_{p}=4.35\times 10^{-4}$

Explanation:

We know, $K_{p}=K_{c}(RT)^{\Delta n}$

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and $\Delta n$ is difference in sum of stoichiometric coefficient of products and reactants

Here $\Delta n=(2)-(2+1)=-1$ and T = 311 K

So, $K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}$

Hence value of equilibrium constant in terms of partial pressure $(K_{p})$ is $4.35\times 10^{-4}$

4 0

3 years ago