Which term best describes an electrical circuit?
Select one:
a. steady flow
b. quick burst
c. short-lived phenomenom
d. dangerous explosion

Answers

Answer 1
Out of all of these, the only one that will ever make sense is A: Steady Flow.
Answer 2
Final answer:

An electrical circuit is best described as a 'steady flow.' Electricity in a circuit flows continuously from the power source, through the device being powered, and back to the source.

Explanation:

The term that best describes an electrical circuit is steady flow. In a basic electrical circuit, electricity moves from a power source like a battery through conductive materials such as wires to the device that it's powering (like a light bulb), and then it returns to the source. This movement is a continuous or steady flow of electric charge, and it usually happens until the power source is depleted or the circuit is otherwise disrupted.

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Related Questions

Which symbol can be used to indicate a chemical reaction that takes place in the presence of a catalyst?
A: 350° C
B: Pt
C: 0.5 atm
D: the triangle

Answers

The correct answer is option B i.e. Pt

Following are the reason for identifying it as a correct answer. 

Option A reflects the temperature at which reaction has been performed. It doesn't signify use of catalyst.
Option B signifies that, Pt is employed as a foreign material in the reaction system. The role of this foreign material is to accelerate the reaction rate. Such material is referred as catalyst. With reference to catalyst, it must be noted that, Pt is a heterogeneous catalyst. 
Option C signifies the pressure of reactant gas, at which reaction is carried out. atm is the unit of pressure of reactant gas. 
Option D is the symbol that is very often used to depict that reaction is performed at higher temperature. 

Thus, among all the available options, option B i.e. Pt is a correct answer 

The symbol that can be used to indicate a chemical reaction that takes place in the presence of a catalyst is letter B: Pt. Platinum is the metal which is often used as a heterogeneous catalysts for several reactions. Catalyst lowers the activation barrier of reaction, thereby accelerating the reactions rates.

Fill in the missing blank 2C4H6 + _______ → 8CO2 + 6H2O

Answers

 the  missing blank is filled by 11 O2
that is
= 2C4H6 + 11O2 = 8 CO2 + 6H2O

C4H6 reacted with oxygen (O2 )through the process of combustion to form carbon iv oxide (CO2)  and water (H2O) . 11 infront  of  O2 is to make sure the molecules of O2 is balanced in both reactant side and the  product side

Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?

Answers

[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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What element is found in all organic compounds?
a.carbon?

Answers

The element that is found in all organic compounds is Carbon

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%?

Answers

m(H₂O) = 255 g.
m₁(NaCl) = 10.0 g.
m₁(solution) = 255 g + 10 g = 265 g.
ω₁ = 10 g / 265 g · 100%.
ω₁ = 3.77% ÷ 100% = 0.0377.
ω₂= 10% ÷ 100% = 0.1.
ω₂= m₁(NaCl) + m₂(NaCl) / m₁(solution) + m₂(NaCl).
0.1 = 10 g + m₂(NaCl) / 265 g + m₂(NaCl).
26.5 g + 0.1·m₂(NaCl) = 10 g + m₂(NaCl).
0.9·m₂(NaCl) = 16.5 g.
m₂(NaCl) = 18.33 g.

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?

Answers

The  grams  of caH2  that  are  needed to  generate 143 L  of H2  is calculated  as follows
by  use of ideal  gas  equation Pv=nRT  where n  is  number  of moles, calculate   the  moles of H2  produced
by making n  the subject  of the formula  
n= PV/RT
p= 827 torr
R(gas  constant)= 62.36 L.torr/mol.K
T= 273 +22=295 k
V=143 L
n  =(827 torr  x143 L)/ 62.36 L.torr/mol.k x295 k) =6.43   moles

write the  reacting equation
that is caH2 +2H2O= Ca(OH)2 + 2H2

by  use of mole  ratio between CaH2  to H2  which is  1 :2 the moles of CaH2 = 6.43 x1/2=3.215  moles

mass of CaH2 is therefore= moles  of CaH2  x molar mass of CaH2

=3.215  moles x 42 g/mol = 135.03  grams

Marshall determines that a gas has a gage pressure of 276 kPa what's the absolute pressure of this gas

Answers

To determine the absolute pressure of this gas, all you need to do is to add the value of atmospheric pressure and the value of gage pressure.

Atmospheric pressure is equivalent to 100 kPa. 
Gage pressure is 276 kPa.

Then, we add both values. 

N = 100 kPa + 276 kPa
N =  376 kPa

The absolute pressure of this gas is 376 kPa.

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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

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

Answers

Boyle's law gives the relationship between pressure and volume of gas. 
It states that at constant temperature, pressure is inversely proportional to volume of gas.
PV = k
where P - pressure, V - volume and k - constant 
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
substituting these values in the equation 
4.00 atm x 500 L = 8.0 atm x V
V = 250 L 
new volume is 250 L

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Lakes that have been acidified by acid rain can be neutralized by limiting the addition of limestone how much limestone is required to completely neutralize a 4.3 billion liter lake with a ph of 5.5

Answers

From the given pH, we calculate the concentration of H+:
     [H+] = 10^-pH = 10^-5.5
We then use the volume to solve for the number of moles of H+:
     moles H+ = 10^-5.5M * 4.3x10^9 L =  13598 moles
From the balanced equation of the neutralization of hydrogen ion by limestone written as
     CaCO3(s) + 2H+(aq)  → Ca2+(aq) + H2CO3(aq)
we use the mole ratio of limestone CaCO3 and H+ from their coefficients, which is 1 mole of CaCO3 is to react with 2 moles of H+, to compute for the mass of the limestone:
     mass CaCO3 = 13598mol H+(1mol CaCO3/2mol H+)
                               (100.0869g CaCO3/1mol CaCO3)(1kg/1000g) 
                            = 680 kg

The amount of limestone required to neutralize the lake is approximately [tex]\( 2.653 \times 10^4 \, \text{kg} \).[/tex]

To determine how much limestone ([tex]CaCO_3[/tex]) is required to neutralize a lake acidified by sulfuric acid ([tex]H_2SO_4[/tex]), we need to follow these steps:

1. Calculate the total mass of [tex]H_2SO_4[/tex] in the lake.

2. Determine the moles of [tex]H_2SO_4[/tex] present.

3. Use stoichiometry to find the moles of [tex]CaCO}_3[/tex] required to neutralize the [tex]H_2SO_4[/tex]

4. Convert the moles of [tex]CaCO}_3[/tex] to mass in kilograms.

Step 1: Calculate the Total Mass of [tex]H_2SO_4[/tex] in the Lake

The lake volume is [tex]\( 5.2 \times 10^9 \) liters.[/tex]

The concentration of [tex]H_2SO_4[/tex] is [tex]\( 5.0 \times 10^{-3} \) g/L.[/tex]

Total mass of [tex]H_2SO_4[/tex]

[tex]\[ \text{Total mass of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} \][/tex]

[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 5.0 \times 10^{-3} \, \text{g/L} \times 5.2 \times 10^9 \, \text{L} \][/tex]

[tex]\[ \text{Total mass of H}_2\text{SO}_4 = 2.6 \times 10^7 \, \text{g} \][/tex]

Step 2: Determine the Moles of [tex]H_2SO_4[/tex]

Molar mass of [tex]H_2SO_4[/tex]

[tex]\[ \text{H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol} \][/tex]

Moles of [tex]H_2SO_4[/tex]

[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Total mass of H}_2\text{SO}_4}{\text{Molar mass of H}_2\text{SO}_4} \][/tex]

[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{2.6 \times 10^7 \, \text{g}}{98 \, \text{g/mol}} \][/tex]

[tex]\[ \text{Moles of H}_2\text{SO}_4 = 2.653 \times 10^5 \, \text{mol} \][/tex]

Step 3: Use Stoichiometry to Find the Moles of [tex]CaCO}_3[/tex] Required

The neutralization reaction between [tex]H_2SO_4[/tex] and [tex]CaCO}_3[/tex] is:

[tex]\[ \text{H}_2\text{SO}_4 + \text{CaCO}_3 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]

From the balanced equation, 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]CaCO}_3[/tex]

Therefore, moles of [tex]CaCO}_3[/tex] required:

[tex]\[ \text{Moles of CaCO}_3 = \text{Moles of H}_2\text{SO}_4 \][/tex]

[tex]\[ \text{Moles of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \][/tex]

Step 4: Convert the Moles of [tex]CaCO}_3[/tex] to Mass in Kilograms

Molar mass of [tex]CaCO}_3[/tex]

[tex]\[ \text{CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol} \][/tex]

Mass of [tex]CaCO}_3[/tex]

[tex]\[ \text{Mass of CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3 \][/tex]

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^5 \, \text{mol} \times 100 \, \text{g/mol} \][/tex]

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \][/tex]

Convert grams to kilograms:

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^7 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} \][/tex]

[tex]\[ \text{Mass of CaCO}_3 = 2.653 \times 10^4 \, \text{kg} \][/tex]

The complete Question is

Lakes that have been acidified by acid rain can be neutralized by the addition of limestone (CaCO3). How much limestone in kg would be required to completely neutralize a 5.2 x 10^9 L lake containing 5.0 x 10-3 g of H2SO4 per liter?

What are the products obtained in the electrolysis of molten nai?

Answers

Final answer:

The electrolysis of molten sodium iodide (NaI) yields sodium metal and iodine gas. The process involves the migration of ions to respective electrodes, gaining or losing electrons, thus getting reduced or oxidized respectively.

Explanation:

The products obtained from the electrolysis of molten sodium iodide (NaI) are sodium (Na) and iodine (I2). During electrolysis, sodium ions are reduced at the cathode to form sodium metal, shown by the half-equation 2Na+ + 2e- → 2Na. Meanwhile, iodide ions get oxidized at the anode to produce iodine gas, as depicted by 2I- → I2 + 2e-.

In a nutshell, the process involves electrolysis of molten sodium iodide using a Downs cell. Positively charged sodium ions migrate to the negatively charged cathode and gain electrons, reducing them to sodium metal. Conversely, the negatively charged iodide ions migrate to the positively charged anode and lose electrons, getting oxidized to iodine gas.

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Final answer:

The electrolysis of molten sodium iodide (NaI) produces sodium (Na) and iodine (I2). In the process, sodium ions migrate to the cathode and are reduced to sodium metal, while iodide ions migrate to the anode and are oxidized to iodine gas.

Explanation:

The process of electrolyzing molten sodium iodide, or NaI, is similar both in concept and execution to the electrolysis of molten sodium chloride, which is a better-known and more commonly discussed process. Electrolysis of molten sodium iodide will produce sodium (Na) and iodine (I) as products. Here's a description of how it works:

In the set-up, you would have a Downs cell. The Downs cell contains molten sodium iodide. The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Iodide ions migrate to the positively charged anode, lose electrons, and undergo oxidation to iodine gas.

The overall cell reaction would be: 2NaI -> 2Na + I2

This is a simple explanation of the process and should suffice for your understanding. Do keep in mind that actual industrial processes could have additional steps and complexities not addressed here.

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Which of the following is not a group found in a protein?
Carboxyl Group
L Group
R Group
Amino Group

Answers

The answer is C: L Group.

Answer:

L Group

Explanation:

Proteins are made up of amino acids, these amino acids determine the protein's functional groups, in the image I annexed you can observe that amino acids have an amino group (NH2), an acid group(COOH) and a R group, a lateral chain that varies depending on the amino acid, in the protein these amino acids are linked through a peptide bond leaving one side of the chain with an acid group and the other side with an amide chain, keeping the R groups as well.

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What is the molarity of a solution which contains 58.4 g of sodium chloride dissolved in of solution?

Answers

Final answer:

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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Which statement about ionic bonds is true?
A. In ionic bonds atoms share electrons to achieve a stable outer shell.
B. Ionic bonds occur between non-metals.
C. Ionic bonds occur between two metals.
D. In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.

Answers

Option D is the correct one.

In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.

Answer:

In ionic bonds one atom accepts electrons from another atom to achieve a stable outer shell.

Explanation:

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How many grams of NAOH are contained within 0.785 moles of NAOH?

Answers

When you do the dimensional analysis for this problem you should end up with the answer 31.4 grams of NAOH.

Answer: The mass of NaOH for given number of moles is 31.4 grams.

Explanation:

To calculate the mass for given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 0.785 moles

Putting values in above equation, we get:

[tex]0.785mol=\frac{\text{Mass of NaOH}}{40g/mol}\\\\\text{Mass of NaOH}=(0.785mol\times 40g/mol)=31.4g[/tex]

Hence, the mass of NaOH for given number of moles is 31.4 grams.

What mass of solute is contained in 25.4 ml of a 1.56 m potassium bromide solution?

Answers

Answer is: mass of potassium bromide is 4.71 grams.
V(KBr) = 25.4 mL ÷ 1000 mL/L = 0.0254 L, volume of solution.
c(KBr) = 1.56 mol/L.
n(KBr) = c(KBr) · V(KBr).
n(KBr) = 1.56 mol/L  0.054 L.
n(KBr) = 0.0396 mol, amount of substance.
m(KBr) = n(KBr) · M(KBr).
m(KBr) = 0.0396 mol · 119 g/mol.
m(KBr) = 4.71 g.
M - molar mass.

Answer : The mass of solute is 4.72 grams.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

or,

[tex]\text{Molarity}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

In this question, the solute is potassium bromide.

Given:

Volume of solution = 25.4 mL

Molarity = 1.56 M

Molar mass of potassium bromide = 119 g/mole

Now put all the given values in this formula, we get:

[tex]1.56M=\frac{\text{Mass of solute}\times 1000}{119g/mole\times 25.4L}[/tex]

[tex]\text{Mass of solute}=4.72g[/tex]

Therefore, the mass of solute is 4.72 grams.

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?

Answers

The activity of the sample is:
R₀ = 4.5 x 10⁻⁶ Ci (3.70 x 10¹⁰ decays/s / 1 Ci)
     = 166500 decays / s

The number of nuclei is:
N₀ = (166500 decays/s) / (5730 yr * 3.154 x 10⁷s /1yr)  = 9.2 x 10⁻⁷

The mass of a ₆C¹⁴ source is:
m = N₀m₀ = (9.2 x 10⁻⁷) * (2.34 x 10⁻²⁶) = 2.16 x 10⁻³² kg

The ka of hypochlorous acid (hclo) is 3.0 x 10-8 at . what is the % ionization of hypochlorous acid in a 0.015 m aqueous solution of hclo at ?

Answers

Answer is: the % ionization of hypochlorous acid is 0.14.

Balanced chemical reaction (dissociation) of an aqueous solution of hypochlorous acid:

HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).

Ka = [H⁺] · [ClO⁻] / [HClO].

[H⁺] is equilibrium concentration of hydrogen cations or protons.

[ClO⁻] is equilibrium concentration of hypochlorite anions.

[HClO] is equilibrium concentration  of hypochlorous acid.

Ka is the acid dissociation constant.

Ka(HClO) = 3.0·10⁻⁸.

c(HClO) = 0.015 M.

Ka(HClO) = α² · c(HClO).

α = √(3.0·10⁻⁸ ÷ 0.015).

α = 0.0014 · 100% = 0.14%.

The percent ionization of the solution is 0.14%.

The equation of the reaction is;

        HClO(aq) ⇄      H^+(aq)   +    ClO^-(aq)

I        0.015                 0                    0

C       -x                       +x                   +x

E   0.015 - x                  x                      x

Now the Ka of the solution is obtained from;

Ka = [ H^+] [ClO^-]/[HClO]

3.0 x 10^-8= x^2/ 0.015 - x  

3.0 x 10^-8(0.015 - x) = x^2

4.5 x 10^-10 - 3.0 x 10^-8x = x^2

x^2 + 3.0 x 10^-8x - 4.5 x 10^-10 = 0

x = 0.000021 M

Percent ionization =  0.000021 M/0.015 M × 100/1

Percent ionization = 0.14%

Hence, the percent ionization of the solution is 0.14%.

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What type(s) of intermolecular forces are expected between ch3conhch3 molecules?

Answers

Hydrogen bonds.
________________
Final answer:

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.

Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m hcl. what was the initial [oh − ]?

Answers

Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L

∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
                                                           = 0.0199 X 16.08 X 10^(-3)
                                                           = 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = 
[tex] \frac{number of moles of [OH-]}{volume of solution (L)} [/tex] = [tex] \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} [/tex] = 0.03199 M

What is the [H+] if the pH of a solution is 7.90

Answers

pH scale is used to determine how acidic, basic or neutral a solution is.
pH is calculated using the hydrogen ion concentration 
pH = -log[H⁺]
if we know the pH we can calculate the H⁺ concentration 
[H⁺] = antilog(-pH)
[H⁺] = 1.26 x 10⁻⁸ M
therefore [H⁺] is 1.26 x 10⁻⁸ M

Answer:

1.26 x 10^-8

Explanation

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Select the single best answer. why is the boiling point of propanamide, ch3ch2conh2, considerably higher than the boiling point of n,n−dimethylformamide, hcon(ch3)2 (213°c vs. 153°c), even though both compounds are isomeric amides?

Answers

The reason for higher boiling point of Propanamide as compared to N,N−dimethylformamide although both are isomers of each other is due to the difference in the intermolecular interactions among their molecules. 

As shown below, propanamide contains two hydrogen atoms directly attached to Nitrogen atom. In this case the hydrogen atoms become partially positive and can form Hydrogen Bond Interactions with the Nitrogen atoms of neighbor propanamide molecule.

Due to the involvement of Hydrogen bonding which is strongest among intermolecular interactions increases the boiling point of propanamide which are absent in N,N−dimethylformamide.
Final answer:

The higher boiling point of propanamide compared to N,N-dimethylformamide is due to the ability of propanamide to form stronger intermolecular hydrogen bonds, requiring more energy to break and hence, raising its boiling point.

Explanation:

The boiling point of propanamide, CH3CH2CONH2, is considerably higher than that of N,N-dimethylformamide, HCON(CH3)2 because propanamide can form robust intermolecular hydrogen bonds within its molecules. These require a significant amount of energy to break, hence increasing its boiling point. However, in N,N-dimethylformamide, hydrogen bonding is inhibited because the hydrogen atoms are bonded to carbon rather than a more electronegative element like nitrogen or oxygen, which would contribute to stronger hydrogen bonding. Therefore, the boiling point of N,N-dimethylformamide is lower.

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Na has an atomic mass of 23.0g and O has an atomic mass of 16.0g. How many grams of Na are needed to completely react with 40.0g of O2?

Answers

The grams of Na that are needed to complete  to react with 40..0 g of O2 is calculated as  below

find the moles of O2 used = mass/molar mass

= 40 g/32g/mol = 1.25  moles

write the reacting equation
4Na+ O2 = 2Na2O

by use of mole ratio between Na to O2 which is 4 :1

the moles of Na = 1.25 x 4 = 5 moles

mass of Na = mass x molar mass

=  5 moles x 23 g /mol= 115  moles

Final answer:

To completely react with 40.0g of O2, 115.0g of Na are required. This calculation uses stoichiometry and the balanced chemical equation 4Na + O2 → 2Na2O, considering the atomic masses of Na and O.

Explanation:

The question relates to how many grams of sodium (Na) are needed to react completely with 40.0 grams of oxygen (O2). This is a stoichiometry problem that involves understanding the molar mass of the reactants and the chemical reaction equation. The first step is to understand the reaction between sodium and oxygen, which typically forms sodium oxide (Na2O). The balanced chemical equation for this reaction is:

4Na + O2 → 2Na2O

From this equation, we can see that 4 moles of Na react with 1 mole of O2 to produce 2 moles of Na2O. Using the given atomic mass of O (16.0 g/mol) and knowing that oxygen is diatomic in its normal state (O2), the molar mass of O2 is 32.0 g/mol. Therefore, 40.0 g of O2 corresponds to 40.0 g / 32.0 g/mol = 1.25 moles of O2.

Since 4 moles of Na are required for every 1 mole of O2, the moles of Na needed is 4 * 1.25 moles = 5 moles of Na. Given the atomic mass of Na as 23.0 g/mol, the mass of Na required is 5 moles * 23.0 g/mol = 115.0 grams. Therefore, to completely react with 40.0 grams of O2, 115.0 grams of Na are needed.

Suppose that 26.89 ml of vinegar solution requires 33.23 ml of the 0.09892 m sodium hydroxide solution to reach the endpoint. calculate the molar concentration of the vinegar solution.

Answers

Number of millimoles of NaOH consumed in present titration =33.23 X 0.09892
                                                                                                 = 3.287

Thus, number of millimoles of vinegar present in sample = 3.287

Now, Molarity of vinegar solution = [tex] \frac{\text{number of millimoles of vinegar}}{\text{volume of solution (ml)}} [/tex]
                                                     = [tex] \frac{3.287}{26.89} [/tex]
                                                     = 0.1222 mol/dm3

Molar concentration of vinegar present in sample solution is 0.1222 M 

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.

Answers

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.

An apple pudding is prepared by combining apples, sugar, butter, and lemon juice. To which category does it belong? compound mixture element molecule atom

Answers

Mixture as...

Each individual ingredient is an element.
Compound means they are chemically bound together which is not true in this case.

mixture   between them  

What are the oxidation numbers of potassium, sulfur, and oxygen, respectively, in k2so4?

Answers

Final answer:

In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) are +1, +6 and -2, respectively.

Explanation:

In the compound K2SO4, the oxidation numbers for potassium (K), sulfur (S) and oxygen (O) can be determined based on established guidelines for assigning oxidation states.

In every stable (neutral) atom, the oxidation number is always zero. Therefore, for potassium, in its ionic form, it has an oxidation number of +1.

In general, the oxidation number of oxygen in its compounds is -2. The compound K2SO4 contains 4 oxygen atoms. Therefore, the total oxidation state contributed by oxygen is -8.

To ensure that the compound is electrically neutral, the total oxidation number should be zero. Hence, for sulfur, you would calculate its oxidation state as follows: total oxidation state of compound = (+1 x 2 for potassium) + oxidation state of sulfur + (-2 x 4 for oxygen) = 0. Solving this equation gives the oxidation state for sulfur as +6.

So, the oxidation numbers for potassium, sulfur and oxygen in K2SO4 are +1, +6 and -2, respectively.

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Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -

Answers

The given complex ion is as follow,

                                              [Ru (CN) (CO)₄]⁻

Where;
            [ ]  =  Coordination Sphere

            Ru  =  Central Metal Atom  =  Ruthenium

            CN  =  Cyanide Ligand

            CO  =  Carbonyl Ligand

The charge on Ru is calculated as follow,

                               Ru + (CN) + (CO)₄  =  -1
Where;
            -1  =  overall charge on sphere

             0  =  Charge on neutral CO

            -1  =  Charge on CN

So, Putting values,


                               Ru + (-1) + (0)₄  =  -1

                               Ru - 1 + 0  =  -1

                               Ru - 1  =  -1

                               Ru  =  -1 + 1

                               Ru  =  0
Result:
          Oxidation state of the metal species in each complex [Ru(CN)(CO)₄]⁻ is zero.

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.

Learn more:

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.

If you ix 20.0 ml of a 3.00m sugar solution with 30.0 ml of a 5.69 m sugar solution, you will end up with a sugar solution of?

Answers

Find mow many moles of sugar you have total

20 mL / 1000 mL x 3 M = .06 moles sugar

30 mL / 1000 mL x 5.69 M = .1707 moles sugar

Total = .2307 moles

Find how many mL of solution you have. 20 mL + 30 mL = 50 mL

1000 mL / 50 mL x .2307 moles = 4.614 M solution

If a person mixes 20.0 ml of a 3.00m sugar solution with 30.0 ml of a 5.69m sugar solution, you will end up with a sugar solution of:

4.614m solution

According to the given question, we can see that there is a mixture of 20 milliliters of 3 moles of sugar solution with 30 milliliters of 5.69 moles of another sugar solution, then we need to find the total sugar solution from the mixtures.

As a result of this, we need to first convert the milliliters to moles.

20 mL/1000 mL x 3 moles = 0.06 moles of sugar

30 mL/ 1000 mL x 5.69 moles = 0.1707 moles of sugar

Please note that we are dividing by 1000 mL to convert to liters.

Now, we add up the two values, 0.06 + 0.1707 = 0.2307 moles

Next, we add the total millimeters we have so far which would be

20 mL + 30 mL = 50 mL

Finally, we would convert the 50 mL to liters before we get our final answer.

1000 mL / 50 mL = 20 L

Now, we multiply 20 L by 0.2307 moles = 4.614 M solution

 

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Nitrogen-13 has a half-life of 10 minutes. how much of a 128 mg sample would remain after 20 minutes?

Answers

The half-life of a radioisotope describes the amount of time it takes for said isotope to decay to one-half the original amount present in the sample.

Nitrogen-13, because it has a half-life of ten minutes, will experience two half-lives over the course of the twenty minute period. This means that 25% of the isotope will remain after this.

0.25 x 128mg = 32mg

32mg of Nitrogen-13 will remain after 20 minutes.
Final answer:

After 20 minutes, approximately 32 mg of the 128 mg sample of Nitrogen-13 would remain.

Explanation:

The half-life of a radioactive isotope is the time required for half of the atoms in a sample to decay.

In this case, we are given that the half-life of Nitrogen-13 is 10 minutes. This means that after every 10 minutes, half of the sample will decay.

Since 20 minutes have passed, we need to determine how many half-lives have occurred. There have been 2 half-lives because 20 divided by 10 equals 2.

Therefore, after 2 half-lives, one-fourth of the sample will remain (since half of the original sample will decay after each half-life).

To find out how much of a 128 mg sample would remain after 20 minutes, we multiply the original amount by one-fourth:

128 mg x 1/4 = 32 mg

After 20 minutes, approximately 32 mg of the 128 mg sample of Nitrogen-13 would remain.

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The reaction 4 a + c + h → d has the mechanism below. what is the rate law? 1) 2 a → b fast 2) 2 b → b2 fast 3) b2 + c → g slow 4) g + h → d fast
a. rate = k[a]2[c]
b. rate = k[a]4[h][c]
c. rate = k[a]2
d. rate = k[b]2[c]
e. rate = k[a]4[c]

Answers

Final answer:

To determine the rate law, the rate-determining step must be identified, which is the third step in the given reaction mechanism. The rate law correlates with the concentration of reactants in this slow step, resulting in a rate law of rate [tex]= k[A]^2[C][/tex] correct answer to the question is a. rate = [tex]k[A]^2[C].[/tex]

Explanation:

To determine the rate law for a reaction with a given mechanism, it is essential to identify the rate-determining step, as the rate law is based on this slowest step. In the provided mechanism, the slow step is the third one, where B2 reacts with C to form G. The overall reaction is 4 A + C + H → D, and the steps are:

2 A → B (fast)

2 B → B2 (fast)

B2 + C → G (slow)

G + H → D (fast)

Because step 3 is the slow, rate-determining step, the rate law will be based on the concentrations of the reactants in this step. Since B is formed from A in a fast step and B2 is formed from 2 B, the rate of formation of B2 is dependent on the concentration of A. However, as B2 forms immediately before the slow step, we look at the concentration of A instead of B2 when writing the rate law.

With the stoichiometry of 2 A forming B and then 2 B forming B2, we can note that the concentration of B (and hence B2) is proportional to [tex][A]^2[/tex]efore, for the rate-determining step, the rate law is:

[tex]rate = k[B2][C] → rate = k[A]^2[C][/tex]

So the correct rate law for the overall reaction is:

[tex]rate = k[A]^2[C][/tex]

[tex]rate = k[A]^2[C][/tex]

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