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 $\begin{array}{c}{141} \\ {58}\end{array} C e.$

Explanation:

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

Describe how the air pressure on Earth affects the weather.

11111nata11111 [884]

Answer:

As the pressure decreases, the amount of oxygen available to breathe also decreases. Atmospheric pressure is an indicator of weather. When a low-pressure system moves into an area, it usually leads to cloudiness, wind, and precipitation. High-pressure systems usually lead to fair, calm weather.

5 0

2 years ago

Dinitrogen pentoxide can be made by the reaction of nitrogen with oxygen. Consider the reaction

Rom4ik [11]

The only reactants are Nitrogen and Oxygen and the only product is dinitrogen pentoxide. Following the law of conservation of mass, the total amount of reactants must equal the total amount of products. Thus, the total mass of dinitrogen pentoxide:
35 + 48
= 83 grams

7 0

2 years ago

Find the number of mol-atom in 7.1 gr cl

marin [14]

I don’t know if this would help but my cousin said this: Molecular weight of chlorine = 71, so 7.1/71 = 0.1 mol of Cl2

7 0

2 years ago

What is the percent composition by mass of oxygen in

KiRa [710]

What is the percent composition by mass of oxygen in magnesium oxide, MgO?

Answer: 39.7 percent

4 0

2 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

For $N_2$ :

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

For $H_2$ :

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

2 years ago