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stiks02 [169]
4 years ago
12

A certain radionuclide is used to diagnose lymphoma. An atom of this radionuclide contains 31 protons and 36 neutrons. Which sta

tement is correct about the radionuclide?
A.The atomic number is 31, the mass number is 67, and its symbol is 67/31 GA.
B.The atomic number is 31, the mass number is 67, and its symbol is 31/64 GA.
C.The atomic number is 67, the mass number is 31, and its symbol is 67/31 GA .
D.The atomic number is 67, the mass number is 31, and its symbol is 31/64 GA .
Chemistry
2 answers:
Tasya [4]4 years ago
5 0
The atomic number is the total number of protons in the nucleus. The mass number is the total number of particles in the nucleus. That includes the neutrons and the protons. So, the atomic number is 31. When you look at the element on the periodic table with a designated atomic number of 31, that would be Gallium, with a symbol of Ga. Mass number is 31 + 36 = 67. So, the answer is A.
zzz [600]4 years ago
3 0

Answer:  A. The atomic number is 31, the mass number is 67, and its symbol is _{31}^{67\textrm{Ga}

Explanation:

Atomic number is defined as the number of protons or number of electrons that are present in an electrically neutral atom.

Atomic number = Number of protons = 31

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons  = 31 + 36 = 67

The isotopic representation of a radionuclide is: _Z^A\textrm{X}

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

Thus the atom is written as _{31}^{67\textrm{Ga}

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A graduated cylinder contains 20.0 mL of water. An irregularly shaped object is placed in
RideAnS [48]

Answer:

7.17 g/mL

Explanation:

Density = Mass / Volume

Given mass of object : 80.4g

We are not given the objects volume, however we are given that when put inside of a container with 20.0mL water, the level of the water rises to 31.2 mL.

31.2 - 20 = 11.2

The water level rose 11.2 mL when the object was placed inside of the cylinder containing the water therefore the object has a volume of 11.2mL

Now to find it's density

recall that density = mass / volume

mass = 80.4 and volume = 11.2mL

so density = 80.4/11.2 = 7.17 g/mL

8 0
2 years ago
A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL
d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

pH = 0.70

Putting values in above equation, we get:

0.70=-\log[H^+]

[H^+]=10^{-0.70}=0.199M

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated nitric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

We are given:

M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL

Hence, the volume of concentrated solution required is 9.95 mL

6 0
3 years ago
A water treatment plant has 4 settling tanks that operate in parallel (the flow gets split into 4 equal flow streams), and each
ale4655 [162]

Answer:

a) When the 4 tanks operate in parallel the retention time is 1.26 hours.  

b) If the tanks are in series, the retention time would be 0.31 hours

Explanation:

The plant has 4 tanks, each tank has a volume V = 600 m_3. The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)

When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.

Firstly, we should convert the MGD to m_3 /day. In that sense, we can calculate the retention time using the tank volume in m_3.

Ft =12 \frac{MG}{day}  (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}

After that, we should divide the total flow by four, because we have four tanks.

F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}

To calculate the retention time we divide the total volume V by the flow of each tank F1.

t1 =\frac{V1}{F1} = \frac{600 m^3}{11340  \frac{m^3}{d} }  = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h

After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.

b)

If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.

In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.

F2 = F0

To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.

t2 =\frac{V1}{F2} = \frac{600 m^3}{45360  \frac{m^3}{d} }  = 0.01 day\\t2 = 0.01  d (\frac{24 h}{1d} ) = 0.31 h

After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.

5 0
3 years ago
What would you expect from an endothermic reaction?
Zanzabum
D. The products will have more energy than the reactants.

6 0
3 years ago
Consider the reaction:
stich3 [128]

Answer:

\large \boxed{\text{-851.4 kJ/mol}}

Explanation:

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})

                            2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹:         0         -824.3         -1675.7         0

\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}

7 0
4 years ago
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