Answer:
Geiger counter and scintillation counter
Explanation:
The Geiger counter was invented by Hans Geiger in 1908 to measure the levels of radiation in bodies and the environment, so it is one of the indispensable equipment for the inspector to detect radiation leaks in a nuclear power plant. It contains a tube with argon, which ionizes by being crossed by alpha and beta particles of radiation, closing the electric circuit and triggering the counter.
Similarly, a scintillation detector is an apparatus used to detect ionizing radiation. When something in the environment has been reached by radiation, this detector emits a small ray of light, indicating the radiation contamination.
Answer:
A. Geiger counter and scintillation counter
Explanation:
When water reaches its boiling point and turns into water vapor what happens to the molecules structure
A gas sample contains 4.00g of CH4 and 2.00g of He. What is the volume of the sample at STP?
To calculate the volume of the gas sample at standard temperature and pressure (STP), we find the moles of each gas (CH4 and He) using their molar masses and then multiply by the molar volume (22.4 L/mol) to get their individual volumes at STP. Adding these volumes together gives us the total volume of the gas sample, which is 16.8 L.
Explanation:To calculate the volume of a gas sample at standard temperature and pressure (STP), we can use the molar volume concept where one mole of any gas occupies 22.4 liters at STP. Methane (CH4) has a molar mass of 16.00 g/mol, and helium (He) has a molar mass of 4.00 g/mol.
To find the moles of CH4, we divide 4.00 g by its molar mass, 16.00 g/mol, which gives us 0.250 moles. Similarly, for He, we divide 2.00 g by its molar mass, 4.00 g/mol, which gives us 0.500 moles. To find the total volume, we sum the volumes of CH4 and He after multiplying their mole quantities by the molar volume (22.4 L/mol).
Total volume of CH4 = 0.250 moles × 22.4 L/mol = 5.60 L
Total volume of He = 0.500 moles × 22.4 L/mol = 11.2 L
Therefore, the total volume of the gas sample at STP is 5.60 L + 11.2 L = 16.8 L.
In the nuclear transmutation represented by 168o(p, \alpha) 137n, the emitted particle is ________. in the nuclear transmutation represented by o(p, \alpha) n, the emitted particle is ________. a beta particle. a proton. an alpha particle. a neutron. a positron.
In the nuclear transmutation represented by 168o(p, \alpha) 137n, the emitted particle is alpha. in the nuclear transmutation represented by o(p, \alpha) n, the emitted particle is alpha.
In nuclear reactions, an alpha particle is a common emitted particle. An alpha particle is a relatively heavy particle compared to others, consisting of two protons and two neutrons. The emission of an alpha particle is a form of nuclear decay or transmutation, and it is frequently observed in certain nuclear reactions involving heavy elements.
The emission of an alpha particle is often associated with nuclear decay or transmutation processes. In these processes, the nucleus of an unstable atom transforms into a more stable one by releasing an alpha particle. This helps reduce the excess energy and stabilize the nucleus.
Alpha decay is more commonly observed in heavy elements, particularly those with a high atomic number. Elements with larger atomic nuclei tend to be less stable, and alpha decay is one way for them to reach a more stable configuration.
What is the mass of a 4.50-μci 146c source? the half-life of 146c is 5730 yr.
express your answer to three significant figures and include the appropriate units?
Vanillin, c8h8o3 (m = 152 g/mol), is the molecule responsible for the vanilla flavor in food. how many oxygen atoms are present in a 45.0 mg sample of vanillin?
In a 45.0 mg sample of vanillin, there are approximately 5.36 x 10^20 oxygen atoms. The calculation involves converting the mass of the sample to grams, calculating the number of moles, and using Avogadro's number to determine the number of oxygen atoms.
Explanation:The question is asking how many oxygen atoms are present in a 45.0 mg sample of vanillin, which is a molecule responsible for the vanilla flavor in food. The molecular formula of vanillin is C8H8O3 and its molar mass is 152 g/mol.
First, we have to convert the mass of the sample from milligrams (mg) to grams (g) because the molar mass is in grams. Therefore, 45.0 mg equals 0.045 g.
Next, we calculate the number of moles of vanillin in the sample using the equation:
Number of moles = Mass / Molar mass
Substituting the given values:
Number of moles = 0.045 g / 152 g/mol = 2.96 x 10^-4 moles.
Each molecule of vanillin contains 3 oxygen atoms. Therefore, in one mole of vanillin, there are 3 moles of oxygen atoms. From the concept of Avogadro's number, we know that one mole of any substance contains 6.02 x 10^23 entities (atoms, molecules, ions etc.).
Therefore, in 2.96 x 10^-4 moles of vanillin, we have (3 moles O atoms / mole of vanillin) x (2.96 x 10^-4 moles of vanillin) x (6.02 x 10^23 O atoms / mole of O atoms) = 5.36 x 10^20 oxygen atoms.
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In a lead-acid storage battery, pbo2 is reduced to
What is the molarity of a solution which contains 58.4 g of sodium chloride dissolved in of solution?
The molarity of a solution is calculated by the formula: Molarity = moles of solute divided by volume of solution in liters. For the mentioned solution, we have roughly 1 mole of sodium chloride. The molarity can be obtained by dividing this number by the volume of the solution in liters.
Explanation:The molarity of a solution is given by the formula: Molarity (M) = moles of solute / volume of solution in liters.
Via the provided information, we know that one mole of sodium chloride (NaCl) weighs 58.44 g, and we have 58.4 g NaCl in our solution, roughly equivalent to 1 mole. The volume of the solution unfortunately is not given in the question. If this missing value can be retrieved, the molarity can be calculated by dividing the number of moles (which is 1 in this case) by the volume of the solution in liters.
Assuming a 1 L solution for simplicity, the molarity will be: 1 mol NaCl / 1 L solution = 1 M NaCl solution.
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Methanol (ch3oh) can be made by the reaction of co with h2: co(g)+2h2(g)⇌ch3oh(g) to maximize the equilibrium yield of methanol, would you use a high or low temperature?
Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
The oxidation state of the Ru metal in the complex [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is [tex]\boxed{{\text{zero}}}[/tex] .
Further explanation:
Oxidation number:
The oxidation number is used to represent the formal charge on an atom. It also shows that gain or loss of electrons by the atom. Oxidation number can be a positive or negative number but cannot be fractional.
The rules to identify the oxidation state:
(1) The oxidation state of an atom in the elemental form is zero.
(2) The total charge on the species is equal to the sum of the oxidation state of individual atoms.
(3) The oxidation state of halides is -1. For example, fluorine, chlorine, bromine, iodine have -1 oxidation state.
(4)Hydrogen has a +1 oxidation state.
(5) Oxygen has an oxidation state -2.
(6) In a coordination compound, neutral ligands have zero oxidation state and negative ligands such as CN have -1 oxidation.
The given compound is [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex]
Here, CN is a negative ligand thus oxidation state is -1 and CO is a neutral ligand thus it has 0 oxidation state. Also, the complex has -1 negative charge.
The expression to calculate the oxidation state in [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is,
[tex]\left[{\left({{\text{oxidation state of Ru}}}\right)+\left({{\text{oxidation state of CN}}}\right)+4\left({{\text{oxidation state of CO}}}\right)}\right]=-1[/tex]
…… (1)
Rearrange equation (1) for the oxidation state of Ru.
[tex]{\text{Oxidation state of Ru}}=\left[{-\left({{\text{oxidation state of CN}}}\right)-4\left({{\text{oxidation state of CO}}}\right)-1}\right][/tex]
…… (2)
Substitute -1for the oxidation [tex]{\text{state}}[/tex] of CN and 0 for the oxidation state of CO in equation (2).
[tex]\begin{aligned}{\text{Oxidation state of Ru}}&=\left[{-\left({-{\text{1}}}\right)-4\left({\text{0}}\right)-1}\right]\\&=0\\\end{aligned}[/tex]
The oxidation state of Ru is zero.
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1. General statement that is not applied on metals: https://brainly.com/question/2474874
2. The neutral element represented by the excited state electronic configuration: https://brainly.com/question/9616334
Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Coordination complex
Keywords: Oxidation state, metal complex, ru(cn)(co)4-, formal charge, cynide, carbonyl, zero, hydrogen, oxygen, 0 and -1.
Fill in the missing blank 2C4H6 + _______ → 8CO2 + 6H2O
What is the effect of removing some so3 from a system initially at equilibrium?
Answer:
Right
Left
Left
Right
Explanation:
For the equilibrium system described by this equation, what will happen if SO3 is removed?
The equilibrium shifts to the
✔ right
What will happen if NO is added?
The equilibrium shifts to the
✔ left
.For the equilibrium system described by this equation, what will happen if SO2 is removed?
The equilibrium shifts to the
✔ left
.
What will happen if NO2 is added?
The equilibrium shifts to the
✔ right
What type(s) of intermolecular forces are expected between ch3conhch3 molecules?
The intermolecular forces between CH3CONHCH3 molecules are dipole-dipole attractions and hydrogen bonding.
Explanation:The type of intermolecular forces that are expected between CH3CONHCH3 molecules are dipole-dipole attractions and hydrogen bonding.
CH3CONHCH3, also known as acetamide, has a polar carbonyl group (C=O) and an amide group (C-NH). The oxygen atom in the carbonyl group is more electronegative than the carbon atom, resulting in a partial negative charge on the oxygen and a partial positive charge on the carbon. This dipole moment allows for dipole-dipole attractions between acetamide molecules.
In addition, the hydrogen atoms in the amide group can form hydrogen bonds with the lone pairs of electrons on the nitrogen atom of neighboring acetamide molecules. Hydrogen bonding is a stronger type of dipole-dipole attraction and is responsible for the high boiling point and other properties of acetamide.
The presence of which magnetic feature best explains why a magnet can act at a distance on other magnets or on objects containing certain metals?
Answer:
Magnetic fields.
Explanation:
Hello,
The presence of a magnetic fields best explains why a permanent magnet can act on another magnetic objects when are separated by a certain distance. There exist magnetic field lines which are emanated from the north pole to the south pole whose path is curved. Now, take into account that the longer the distance between the objects, the lower the magnetic field or force.
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What is the final step in the scientific method
Magnesium hydroxide (Mg(OH)2): g/mol Iron(III) oxide (Fe2O3): g/mol
Answer:
Magnesium hydroxide (Mg(OH)2): 58.33 g/mol
Iron(III) oxide (Fe2O3): 159.70 g/mol
Explanation:
To find the molar masses of both magnesium hydroxide and iron(III) oxide, add each molar mass in each compound.
Solving:
[tex]\section*{Molar Mass of Magnesium Hydroxide (Mg(OH)_2):}\textbf{Identify Atomic Masses}\begin{itemize} \item Magnesium (Mg): 24.31 g/mol \item Oxygen (O): 16.00 g/mol (2 oxygen atoms) \item Hydrogen (H): 1.01 g/mol (2 hydrogen atoms)\end{itemize}[/tex]
[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Mg(OH)}_2 = \text{Mg} + 2 \times (\text{O} + \text{H})\]\[= 24.31 \, \text{g/mol} + 2 \times (16.00 \, \text{g/mol} + 1.01 \, \text{g/mol})\]\\\[= 24.31 \, \text{g/mol} + 2 \times 17.01 \, \text{g/mol}\]\[= 24.31 \, \text{g/mol} + 34.02 \, \text{g/mol} = \boxed{58.33 \, \text{g/mol}}\][/tex]
[tex]\hrulefill[/tex]
[tex]\section*{Molar Mass of Iron(III) Oxide (Fe_2\text{O}_3):}\textbf{Identify Atomic Masses}\begin{itemize} \item Iron (Fe): 55.85 g/mol (2 iron atoms) \item Oxygen (O): 16.00 g/mol (3 oxygen atoms)\end{itemize}[/tex]
[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Fe}_2\text{O}_3 = 2 \times \text{Fe} + 3 \times \text{O}\]\\\[= 2 \times 55.85 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol}\]\\\[= 111.70 \, \text{g/mol} + 48.00 \, \text{g/mol} = \boxed{159.70 \, \text{g/mol}}\][/tex]
[tex]\hrulefill[/tex]
Marshall determines that a gas has a gage pressure of 276 kPa what's the absolute pressure of this gas
And object has a mass of 120 KG’s on the moon what is the force of gravity acting on the object of the moon
Carbon tetrachloride will have which shape?
A) bent
B) linear
C) tetrahedral
D) triangular
What is the mass occupied by 44.8 l of nitrogen (n2) at standard conditions?
Click to select the correct answers. Click again to unselect answers. Leave the incorrect answers unselected.
Nuclear fission and fusion both affect the nucleus of an atom. Choose all of the items below that are correct.
The final products of fission and fusion are elements that are different than the original
Fusion involves the splitting apart of large atoms into smaller ones
Fission occurs mostly with elements heavier than lead on the periodic table
Two hydrogen atoms can go through fission to form helium
Fusion and fission involve a loss or gain of electrons
The answer options about a nuclear reaction which are correct include the following:
A. Nuclear fission and fusion both affect the nucleus of an atom.
B. The final products of fission and fusion are elements that are different than the original.
D. Fission occurs mostly with elements heavier than lead on the periodic table.
A nuclear reaction refers to a reaction in which the nucleus of an atom is transformed or transmuted by either joining (fusion) or splitting (fission) the nucleus of another atom of a radioactive element. Also, a nuclear reaction is always accompanied by a release of energy.
Generally, the two (2) main types of nuclear reaction are:
Nuclear fusion: it involves the joining of two smaller nuclei of atoms to form a single massive or heavier nucleus with the release of energy. Nuclear fission: it involves the collision of a heavy atomic nucleus with a neutron, thereby causing a split and release of energy.From the above, we can deduce the following about a nuclear reaction:
I. The nucleus of an atom is affected by both nuclear fission and fusion.
II. The final products of both nuclear fission and fusion are chemical elements that are different than the original radioactive chemical elements.
III. Nuclear fission occurs mostly with chemical elements heavier than lead (Pb) on the periodic table because the nuclear force holding their atoms together is lesser than the electromagnetic force pushing the nucleus apart.
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Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0 g of fluorine gas, the maximum mass of hydrogen fluoride that could be produced is
How many structural and geometrical isomers are there of chloropropene?
Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?
[tex]\boxed{{\text{ClO,ClO}}_{\text{2}}^ - {\text{,ClO}}_{\text{3}}^ - {\text{,ClO}}_{\text{4}}^ - }[/tex] does not follow the octet rule.
Further Explanation:
Octet rule: states that for the stability of any element it must have a valence shell of eight electrons (octet means a group of eight). This rule is given by Kossel and Lewis.
Atoms of same or different elements with an incomplete electronic configuration that is having electrons less than 8 are unstable and they combine together to form stable molecules with a complete octet. They can combine by sharing of electrons or loss or gain of electrons.
In [tex]{\mathbf{ClO}}[/tex], Chlorine (Cl) has 7 electrons in its outer shell and oxygen has 6 electrons in its outer shell. So chlorine and oxygen share 2 electrons each to complete their octet and attain stability. Chlorine atom in the compound has ten electrons in its outer shell so it doesn’t follow the octet rule.
In [tex]{\mathbf{Cl}}{{\mathbf{O}}^ - }[/tex], Chlorine(Cl) has 7 electrons in its outer shell and oxygen has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] so, it has 7 electrons also in its outer shell. So chlorine and oxygen share one electron each to complete their octet and attain stability. All the atoms have eight electrons in their outer shell so they follow the octet rule.
In [tex]{\mathbf{ClO}}_2^ -[/tex], Chlorine (Cl) has 7 electrons in its outer shell and there is 2 oxygen attached to Cl. One has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] and the other is neutral. So, chlorine and neutral oxygen share 2 electrons each to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it forming a single bond with it. A chlorine atom has ten electrons in its outer shell so the compound doesn’t follow the octet rule.
In [tex]{\mathbf{ClO}}_3^ -[/tex], Chlorine (Cl) has 7 electrons in its outer shell and there is 3 oxygen attached to Cl. One has a unit negative charge on it [tex]\left( {{{\text{O}}^ - }} \right)[/tex] and the other two are neutral. So, chlorine and both neutral oxygen share 2 electrons to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it, forming a single bond with it. Here, Cl has ten electrons in its outer shell, so it doesn’t follow the octet rule.
In [tex]{\mathbf{ClO}}_4^ -[/tex] , Chlorine (Cl) has 7 electrons in its outer shell and there is 4 oxygen attached to Cl. One has a unit negative charge [tex]\left( {{{\text{O}}^ - }} \right)[/tex]on it and the other three are neutral. So, chlorine and both neutral oxygen share 2 electrons to get a stable octet and form a double bond. Similarly, chlorine shares 1 electron with oxygen ion with a negative charge on it, forming a single bond with it.
Here, Cl has fourteen electrons in its outer shell, so it doesn’t follow the octet rule.
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Answer details:
Grade: Secondary School
Subject: Chemistry
Chapter: Chemical Bonding
Keywords: octet rule, stability, oxygen, chlorine, covalent, sharing, ClO2-, ClO3-, ClO4-, ClO-and ClO.
The species that do not obey the octet rule are; ClO, ClO2^-, ClO3^-, ClO4^-.
The octet rule states that atoms must have eight electrons in their outermost shell in order to attain stability. Hence, stable molecules, ions and atoms are expected to contain atoms that obey the octet rule.
However, in some chemical species, atoms of elements do not obey the octet rule. For instance, in ClO, chlorine has seven valence electrons and oxygen has six electrons. The octet rule is clearly violated in this case.
Again, chlorine is able to expand its octet (contain more than eight valence electrons). This is seen in the ions; ClO2^- and ClO3^- in which chlorine contains ten valence electrons and ClO4^- in which chlorine contains fourteen valence electrons.
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Calculate the molarity of a solution that contains 0.175 mol of zncl2 n exactly 150 ml of solution.
Explanation:
Molarity is the number of moles present in liter of a solution.
Mathematically, Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
It is given that no. of moles present into the solution are 0.175 mol and volume is 150 ml.
As 1 ml = 0.001 L. So, 150 ml will be equal to 0.15 L.
Hence, calculate the molarity as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.175 mol}{0.15 L}[/tex]
= 1.16 M
Thus, we can conclude that molarity of the given solution is 1.16 M.
If 255 g of water has 10.0 g of NaCl dissolved into initially, how much NaCl must be added in order to raise the mass percent of NaCl by 10%?
The electron dot structure for CI is
A. Cl (w/ a dot above the C)
B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1)
C. Cl (w/ 2 dots above and to the left of the C, 1 underneath, and 2 dots to the right of the 1)
D. Cl (w/ 2 dots above, to the left of, and under the C, as well as 2 dots to the right of the 1)
The electron dot structure for Cl is ... (P.S. The l in Cl is a lowercase "L", for Chlorine, in case anyone else may have been confused.)
The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1).
Explanation:The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1). The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons.
How many grams of cah2 are needed to generate 143 l of h2 gas if the pressure of h2 is 827 torr at 22 ∘c? g.com?
What element is found in all organic compounds?
a.carbon?
Identify the lowest energy lewis structure for nitrogen oxide
The voltage generated by the zinc concentration cell described by the line notation zn(s) ∣∣ zn2+(aq,0.100 m) ∥∥ zn2+(aq,? m) ∣∣ zn(s) is 14.0 mv at 25 °c. calculate the concentration of the zn2+(aq) ion at the cathode.
Answer: The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}(0.100M,aq.)+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}(?M,aq.)+2e^-\rightarrow Zn(s)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = 14.0 mV = 0.014 V (Conversion factor: 1 V = 1000 mV)
[tex][Zn^{2+}]_{anode}[/tex] = 0.100 M
[tex][Zn^{2+}]_{cathode}[/tex] = ? M
Putting values in above equation, we get:
[tex]0.014=0-\frac{0.0592}{2}\log \frac{0.100M}{[Zn^{2+}]_{cathode}}[/tex]
[tex][Zn^{2+}]_{cathode}=0.295M[/tex]
Hence, the concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
The concentration of Zn²⁺ ion at cathode is 0.295 M
Redox half equationsThe redox half equations are those of oxidation and reduction
Oxidation half reaction: Zn(s) ---> Zn²⁺ + 2 e⁻ [Zn⁺] = 0.100 M
reduction half reaction: Zn²⁺ (aq) + 2 e⁻ ----> Zn (s) [Zn⁺] = y
The Ecell = 14.0mV = 0.014 V
Using the Nernst equation:
Ecell = E°cell - 0.0592/n * log([Zn⁺]anode/[Zn⁺]cathode)since it is a concentration cell, E°cell = 0where n is number moles; n = 2
0.014 = 0 - 0.0592/2 * log (0.1/y)
y = 0.295 M
Therefore, the concentration of Zn²⁺ ion at cathode is 0.295 M
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A gas sample of argon, maintained at constant temperature, occupies a volume of 500. l at 4.00 atm. what is the new volume if the pressure were charged to 8.0 atm
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