The complete combustion of which of the following substances produces carbon dioxide and water?



-CO


-Na2CO3


- MgCl2


-C4H10

Answers

Answer 1
Answer:
            The complete combustion of CH₁₀ produces carbon dioxide and water.

Explanation:
                   Hydrocarbons when heated in the presence of Oxygen produces CO₂ and H₂O along with heat. The reaction of Butane with oxygen is as follow,

                            C₄H₁₀  +  5 O₂    →    4 CO₂  +  2 H₂O

So, 1 mole of Butane reacts with 5 moles of oxygen and produces 4 moles of Carbon dioxide and 2 moles of water along with heat.
Answer 2
D. C4H10 Is your answer. Hope I helped! :)

Related Questions

an aqueous solution of NaOh is used as a drain cleaner.If the concentration of OH^- ions in this solution is 1×10^-5,the concentration of H^+ ions in it would be?

Answers

[OH⁺][OH⁻}= 10 ⁻¹⁴
1*10⁻⁵*[H⁺]=10⁻¹⁴
[H⁺]= 10⁻¹⁴/10⁻⁵=10⁻⁹

[H⁺] = 10⁻⁹ M

What is the value of n when the empirical formula is c3h5 and the molecular mass is 205.4?

Answers

empirical formula is the simplest ratio of whole numbers of components in a compound.
the empirical formula - C₃H₅
mass of empirical formula - (12 g/mol x 3) + (1 g/mol x 5) = 41 g
we have to find how many empirical units make up the molecular formula 
number of empirical units = molecular mass / mass of empirical formula 
empirical units = 205.4 g/mol / 41 g = 5.00
molecular formula - n (C₃H₅)
n - number of empirical units 
therefore n = 5 

"if the ph of a solution is _____ the solution is basic.
a. 2b. 5c. 7d. 10"

Answers

D is correct answer.

If the pH of a solution is should be 10 is the solution is basic.

Hope it helped you.

-Charlie
I'm pretty sure it would be D. :D

which has not been a major source of CFCs

Answers

Answer : Any natural sources of CFC's are not known only the major sources like aerosols, propellants, refrigerants,etc are known. So, if any natural sources are given then it cannot be called as a major source for emitting CFC into environment.

Answer:

• televisions

How much heat energy would be released if 78.1 g of water at 0.00 °c were converted to ice at −57.1 °c. give your answer as a positive number in kilojoules (kj)?

Answers

To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g; 
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C 
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348 
 = 26250.0348 J or 26.250 kJ

Question 10 unsaved in a chemical reaction, the mass of the products question 10 options: has no relationship to the mass of the reactants. is less than the mass of the reactants. is greater than the mass of the reactants. is equal to the mass of the reactants.

Answers

what are youre answer choices????

What is the net ionic equation of 2h+ + so42- + ca2+ + 2i- caso4 + 2h+ + 2i-?

Answers

what  is  the  net  ionic equation
2 H^+  + SO4^2-  + Ca^2+  +  2 i^- = CaSo4  + 2H^+  +2i ^-
cancel  the spectator  ion that is the  ions  which  does  not take  place  nin  the  reaction

for this  case  is  2 H^+  and 2 i^-

the net ionic equation  is  therefore
=Ca^2+  + SO4^2- = CaSO4 

he is right he is correct


When 0.440 mol of aluminum are allowed to react with an excess of chlorine gas, cl2, how many moles of aluminum chloride are produced? when 0.440 mol of aluminum are allowed to react with an excess of chlorine gas, cl2, how many moles of aluminum chloride are produced? 0.880 mol 0.440 mol 0.330 mol 0.220 mol 0.110 mol?

Answers

Final answer:

When 0.440 mol of aluminum react with excess chlorine gas, 0.440 mol of aluminum chloride are produced according to the 1:1 stoichiometric ratio in the balanced chemical equation.

Explanation:

When 0.440 mol of aluminum are allowed to react with an excess of chlorine gas (Cl2), to determine how many moles of aluminum chloride (AlCl3) are produced, we look at the stoichiometry of the balanced equation:

2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)

This reaction shows that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride. Given the stoichiometry is a 1:1 ratio between aluminum and aluminum chloride, if you start with 0.440 mol of aluminum, you would end up producing 0.440 mol of aluminum chloride as well, provided chlorine is in excess.

How many grams of copper Cu, 63.54 g/mol are in 3.42 moles copper

Answers

Answer:
              Mass  =  217.30 g

Solution:

Mass and moles are related to each other as,

                                  Moles  =  Mass / M.mass  ---- (1)

Where;
            Moles  =  3.42

            M.Mass  =  63.54 g.mol⁻¹

Solving eq. 1 for Mass,

                                  Mass  =  Moles × M.mass
Putting Values,
                                  Mass  =  3.42 mol × 63.54 g.mol⁻¹

                                  Mass  =  217.30 g

Ac sample of octane that has a mass of 0.750 g is burned in a bone calorimeter. As a result, the temperature if the calorimeter increases from 21.0 °C to 41°C . The specific heatv of the calorimeter is 1.50 J/(g*°C), and it's nass is 1.0 kg. How much heat is released during the combustion

Answers

How much heat is released during the combustion?

30.0 kJ

Hope this helps. 
From the specific heat of the calorimete which is 1.50 J/(g*°C), and it's mass is 1.0 kg, then the amount of energy  that is released during the combustion is 30.0 kJ. 

The radioactive element​ carbon-14 has a​ half-life of 5750 years. a scientist determined that the bones from a mastodon had lost 78.5​% of their​ carbon-14. how old were the bones at the time they were​ discovered?

Answers

Radioactive elements obey 1st order  kinetics,

For 1st order reaction, relation between rate constant (k) and half life [t(1/2)] is,
k = [tex] \frac{0.693}{t(1/2)} = \frac{0.693}{5750} = 1.205 X 10^-^4 hr^-^1 [/tex]

Also, for 1st order reaction, we have
t = [tex] \frac{2.303}{k} log \frac{\text{initial conc.}}{\text{final conc.}} [/tex]
 
Given that: the bones from a mastodon had lost 78.5​% of their​ C14,
∴ initial conc. of C14 = 100%, conc. of C14 left after time 't' = 21.5%

∴t = [tex] \frac{2.303}{1.205 X 10^(-4)} log \frac{\text{100}}{\text{21.5}} [/tex] = 1.2758 X 10^4 hours

If the number of moles of a gas initially contained in a 2.10 l vessel is doubled, what is the final volume of the gas in liters? (assume the pressure and temperature remain constant.) none of the above 8.40 4.20 1.05 6.30

Answers

Avagadro's law gives the relationship between volume and number of moles of gas. 
At constant temperature and pressure , volume of gas is directly proportional to number of moles of gas. 
V / n = k
where V - volume, n - number of moles , k - constant 
[tex] \frac{V1}{n1} = \frac{V2}{n2} [/tex]
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 

initial number of moles - N
after doubling - 2N
initial volume - 2.10 L
final volume - V
substituting in the equation 

[tex] \frac{2.10L}{N} = \frac{V}{2N} [/tex]
V = 4.20 L
new volume is 4.20 L

A 4.305-g sample of a nonelectrolyte is dissolved in 105 g of water. the solution freezes at -1.23c. calculate the molar mass of the solute. kf for water = 1.86c/m.

Answers

The answer is 62.00 g/mol
Solution: 
Knowing that the freezing point of water is 0°C, temperature change Δt is 
     Δt = 0C - (-1.23°C) = 1.23°C 
Since the van 't Hoff factor i is essentially 1 for non-electrolytes dissolved in water, we calculate for the number of moles x of the compound dissolved from the equation 
     Δt = i Kf m 
     1.23°C = (1) (1.86°C kg mol-1) (x / 0.105 kg) 
     x = 0.069435 mol 
Therefore, the molar mass of the solute is  
     molar mass = 4.305g / 0.069435mol = 62.00 g/mol
Final answer:

The molar mass of the solute is calculated by first finding the molality of the solution using the freezing point depression, then using this to find the number of moles of solute, and finally dividing the mass of the solute by the number of moles. The computed molar mass of the solute is 62.15 g/mol.

Explanation:

To calculate the molar mass of the solute, we first need to calculate the molality of the solution using the freezing point depression formula, ΔTf = iKfm, where i is the van't Hoff factor (which is 1 for a non-electrolyte), Kf is the freezing point depression constant for water (1.86°C/m), and m is molality. We know that ΔTf = -1.23°C. So, molality can be determined as follows: m = ΔTf / (i x Kf) = -1.23°C / (1 x 1.86°C/m) = -0.66mol/kg.

Molality is defined as moles of solute per kilogram of solvent. Therefore, if we multiply the molality by the mass of the solvent (105g or 0.105kg), we get the moles of solute. Moles of solute = m x mass of solvent = -0.66mol/kg x 0.105kg = -0.0693mol.

Finally, to determine the molar mass of the solute, we divide the mass of the solute by the number of moles. Molar mass = mass of solute / moles of solute = 4.305g / -0.0693mol = -62.15g/mol. The negative sign indicates a freezing point depression, but the molar mass of a substance cannot be negative. Therefore, the molar mass of the solute is 62.15g/mol.

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216 j of energy is required to raise the temperature of aluminum from 15o to 35oc. calculate the mass of aluminum. (specific heat capacity of aluminum is 0.90 joc-1g-1).

Answers

energy is supplied to aluminium to raise the temperature of aluminium.
we can use the following equation to calculate the mass of aluminium
H = mcΔt
H - heat energy supplied 
m - mass of material 
c - specific heat capacity of aluminium
Δt - change in temperature - 35 °C - 15 °C = 20 °C
substituting the values in the equation 
216 J = m x 0.90 J°C⁻¹g⁻¹ x 20 °C
m = 12 g
mass of aluminium is 12 g

The mass of aluminum, the heat energy equation Q = mcΔT is used with the given values, resulting in a calculated mass of 12 grams.

The mass of aluminum using the given energy, temperature change, and specific heat capacity, we employ the formula for heat transfer: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Given that Q = 216 J, c = 0.90 J/°C·g, and ΔT = (35 - 15)°C = 20°C, we can rearrange the equation to solve for m:

m = Q / (cΔT) = 216 J / (0.90 J/°C·g · 20°C) = 216 J / (18 J/g) = 12 g

Therefore, the mass of aluminum is 12 grams.

What is a functional group? A-a group of organs in the body that work together to perform a certain function B-a group of living things that are made up of organic compounds C-a group of atoms bonded to carbon that determines how the molecule will react D-a group of organic molecules that work together to speed up reactions

Answers

C-a group of atoms bonded to carbon that determines how the molecule will react

The answer to your question is C.

A solution consists of two parts. one part is the substance that is dissolved. what is the name of this part of a solution?

Answers

Answer:  "solute".
______________________________________________________
The substance that is dissolved in the solution is the "solute".
______________________________________________________
{Note of interest:  In an aqueous solution, the medium in which the solute is dissolved is the "solvent". }
______________________________________________________

An equilibrium mixture of so2, o2, and so3 gases is determined to consist of 2 mol/l so2, 1 mol/l o2, and 4 mol/l so3. what is the equilibrium constant for the system at this temperature? the balanced equation for this reaction is

Answers

The all substances that participate in the reaction are in gaseous phase and the given data are the concentrations when the reaction at equilibrium. Hence, we can use that given concentrations directly to find out the equilibrium constant, Kc.

The balanced equation for the equilibrium is,

                 2SO(g) + O(g) 2SO(g)

Kc = [SO₃(g)]² / [SO₂(g)]² [O₂(g)]
Kc = (4 mol/L)² / (2 mol/L)² (1 mol/L)
Kc = 4 mol⁻¹ L

Hence, the equilibrium constant is 4 mol¹ L

A 98.0°C piece of cadmium (c=.850J/g°C) is placed in 150.0g of 37.0°C water. After sitting for a few minutes, both have a temperature of 38.6°C. What was the mass of the cadmium sample?

Answers

The answer is 19.9 grams cadmium. 
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
     heat lost by cadmium = heat gained by the water
     -qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
     -qcadmium = qwater
     -(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
     mcadmium = 19.9 grams

If 355 mL of 1.50 M aluminum nitrate is added to an excess of sodium sulfate, how many grams of aluminum sulfate will be produced?

Answers

Answer is: 91.1 grams of aluminum sulfate.
Balanced chemical reaction: 2Al(NO₃)₃ + 3Na₂SO₄→ Al₂(SO₄)₃ + 6NaNO₃.
V(Al(NO₃)₃) = 355 mL ÷ 1000 mL/L = 0.355 L.
c(Al(NO₃)₃) = 1.5 mol/L.
n(Al(NO₃)₃) = V(Al(NO₃)₃) · c(Al(NO₃)₃).
n(Al(NO₃)₃) = 0,355 L · 1.5 mol/L.
n(Al(NO₃)₃) = 0.5325 mol.
From chemical reaction: n(Al(NO₃)₃) : n(Al₂(SO₄)₃) = 2 : 1.
n(Al₂(SO₄)₃) = 0.266 mol.
m(Al₂(SO₄)₃) = 0.266 mol · 342.15 g/mol.
m(Al₂(SO₄)₃) = 91.1 g.

What mass of magnesium bromide would be required to prepare 720?

Answers

Complete Question:
                               What mass of magnesium bromide would be required to prepare 720mL of a 0.0939M aqueous solution?

Answer:
            Molarity is calculated using following formula,

                               Molarity  =  Moles / Volume of Solution  ---- (1)

Data Give;
                 Molarity  =  0.0939 mol.L⁻¹

                 Volume  =  720 mL  =  0.72 L

Solving eq.1 for Moles,

                            Moles  =  Molarity × Volume

                            Moles  =  0.0939 mol.L⁻¹ × 0.72 L

                            Moles  =  0.0676 moles

Now convert Moles into Mass using following formula,

                            Moles  =  Mass / M.Mass

Solving for Mass,

                            Mass  =  Moles × M.Mass

                            Mass  =  0.0676 mol × 184.113 g.mol⁻¹

                            Mass  =  12.44 grams

A cube of iron (Cp = 0.450 J/g•°C) with a mass of 55.8 g is heated from 25.0°C to 49.0°C. How much heat is required for this process?

Answers

Q=c(p)*m*Δt
c(p)=0.450 J/g•°C
m= 55.8 g
Δt = t(final)-t(initial)=49.0⁰-25.0⁰ = 24.0⁰C

Q=c(p)*m*Δt = 0.450 J/g•°C *  55.8 g*24.0⁰C ≈ 603 J
Q≈ 603 J

The process in which sediment is dropped and comes to rest is called

Answers

Deposistion is Correct i'm 100% sure 

Answer : The process in which sediment is dropped and comes to rest is called, Deposition.

Explanation:

Deposition : It is a type of process in which the molecules are settle down at the bottom of the container. In this process, there is no intermolecular space present between the atoms of molecule.

This process results in the formation of the sedimentary rocks which are made out of ice, wind and water flows which carry particles in the suspension.

Hence, the process in which sediment is dropped and comes to rest is called, Deposition.

What test or tests do you use to find NaCl

Answers

Sodium chloride, regular table salt, is also known as the mineral halite. The diagram to the right shows how sodium and chlorine atoms pack tightly together to form cube-like units of the compound NaCl. Crystals of table salt imitate this structure-they're shaped like little cubes. You can check this out for yourself by viewing a few grains of salt through a magnifying lens or microscope.

Answer:

Sodium chloride, regular table salt, is also known as the mineral halite. The diagram to the right shows how sodium and chlorine atoms pack tightly together to form cube-like units of the compound NaCl

Explanation:

4. The diagram shows an electrochemical cell with a gold strip (left) and aluminum glasses (right). In your response, do the following: • Label all parts. • Label the cathode and the anode, including the charge on each. • Show the flow of electrons. • Describe what type of electrochemical cell is pictured and what its use is. • Explain how the cell works. Include the oxidation and reduction half-reactions in your explanation.

Answers

The electrochemical cell in the diagram is an electrolytic cell used for gold plating. Electroplating is an electrolytic process used to coat metal objects with a more expensive and less reactive metal. The metal to be plated is used as the anode and the electrolyte solution contains the ions of the metal to be plated.

In the diagram, the gold strip on the left is connected to the positive terminal of the battery and is anode. The aluminum glasses on the right is connected to negative terminal of the battery and is cathode. Anode is positive electrode and cathode is negative.

When the power is switched on, the electrolyte solution splits into ions. Positively charged Au³⁺ ions are attracted to the negatively charged cathode, aluminum glasses and slowly deposit on it. The negatively charged ions are attracted to positively charged anode, the gold strip and release electrons. The electrons move through battery towards the cathode.

Au goes into the solution at the anode and get reduced at the cathode, and is thus plated. A constant concentration of Au³⁺ is maintained in the electrolyte solution surrounding the electrodes.

Anode: Au(s) → Au³⁺(aq) + 3e-​ ............(oxidation)

Cathode: Au³⁺(aq) + 3e​- → Au(s) .........(reduction)

What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 mL of water?

Answers

we are going to use this equation:

ΔT = - i m Kf

when m is the molality of a solution 

i = 2

and ΔT is the change in melting point = T2- 0 °C

and Kf is cryoscopic constant = 1.86C/m

now we need to calculate the molality so we have to get the moles of NaCl first:

moles of NaCl = mass / molar mass

                         = 3.5 g / 58.44 

                         = 0.0599 moles


when the density of water = 1 g / mL and the volume =230 L

∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg 

now we can get the molality = moles NaCl / Kg water

                                                =0.0599moles/0.23Kg

                                                = 0.26 m

∴T2-0 =  - 2 * 0.26 *1.86

∴T2 = -0.967 °C

Answer:

0.952 °C

Explanation:

The change in melting point is computed as:

ΔT = k*m*i

where ΔT is the difference between the melting point of water and of solution, k is a constant (1.86 °C*kg/mol for water), i is the van't Hoff factor (equal to 2 for sodium chloride because 2 ions are obtained after its dissolution), and m is the molality of the solution.

Molar mass of sodium chloride: 58.44 g/mol

Moles of of sodium chloride: mass / molar mass 3.5/58.44 = 0.059 mol

Density of water 1 kg/L

230 mL of water are equivalent to 0.23 L

mass of water: density * volume = 1*0.23 = 0.23 kg

Molality of the solution: m = moles of solute/ kg of solvent = 0.059/0.23 = 0.256

Finally:

ΔT = 1.86*0.256*2 = 0.952 °C

Water melting point: 0 °C

So, the solution melting point is: 0 - 0.952 = 0.952 °C

If sodium loses 1 electron, it will have the same number of valence electrons as which noble gas?

Answers

atomic number of Na -11
 electronic configuration of Na - 2,8,1
outermost shell of Na has one electron. To become stable and to have a complete octet in the outermost shell, it loses that electron in the outermost shell and becomes positively charged.
Then electronic configuration is 2,8 with 10 electrons . Noble gas with atomic number 10 is Neon.Therefore it will have the same number of valence electrons as B) Neon

The mass percent of oxygen in pure glucose, c6h12o6 is 53.3 percent. a chemist analyzes a sample of glucose that contains impurities and determines that the mass percent of oxygen is 49.7 percent. which of the follow impurities could account for the low mass percent of oxygen in the sample?

Answers

Methanol is the most plausible candidate among these potential contaminants because its oxygen content is closest to the reported mass percent of oxygen in the sample (49.7%).

The proportional amount of a certain component (typically a substance or element) in a combination, compound, or solution is expressed as mass percent (also known as weight percent or weightage). It is a percentage that represents the component's mass in relation to the overall mass of the combination, compound, or solution multiplied by 100.

molecular formula mass of [tex]\rm C_6H_{12}O_6[/tex] = 180.18 g/mol

mass percent of oxygen is

(96.00 g/mol / 180.18 g/mol) × 100%

≈ 53.3%

In water, the mass percent of oxygen is:

(16.00 g/mol / 18.02 g/mol) × 100%

≈ 88.9%

In formaldehyde, the mass percent of oxygen is:

(16.00 g/mol / 30.03 g/mol) × 100%

≈ 53.3%

In methanol, the mass percent of oxygen is:

(16.00 g/mol / 32.04 g/mol) × 100%

50.0%

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How much energy is needed to convert 64.0 grams of ice at 0.00?

Answers

Final answer:

Approximately 21.4 kJ of energy is needed to convert 64.0 grams of ice at 0.00°C to liquid water at the same temperature. This is calculated using the concept of 'enthalpy of fusion', which is the energy required to convert a substance from solid to liquid without changing its temperature.

Explanation:

To find out how much energy is needed to convert 64.0 grams of ice at 0.00°C to water at the same temperature, we should first understand that the melting of ice is an endothermic process, meaning it requires energy. The energy required to convert a substance from solid to liquid state without changing its temperature is termed the enthalpy of fusion.

For water, the enthalpy of fusion is 334 kJ/kg. However, our given mass is in grams, so we need to convert it to kilograms, thus we have 0.064 kg of ice. Using the formula Q=mLf (where Q is energy, m is mass, and Lf is the enthalpy of fusion), we can calculate the necessary energy: Q = (0.064 kg)(334 kJ/kg) = 21.376 kJ.

So, approximatley 21.4 kJ of energy is required to convert 64.0 grams of ice at 0.00°C into water at the same temperature.

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what is ½O2 ? pls help me!!!

Answers

Basically,1/2O2 is just one oxygen atom out of the two.the double bind is broken,and you get to oxygen ions instead of oxygen gas.Notably oxygen is diatomic gas,which means that has two oxygen in one molecule.
it should be just O.

In the word equation, sodium oxide water sodium hydroxide, the formula for sodium hydroxide is represented by

Answers

Answer:
            Formula for sodium hydroxide is represented by NaOH.

Explanation:
                     The reaction of Sodium oxide with water is as follow,

                               Na₂O  +  H₂O    →    2 NaOH

Na₂O is considered as a strong basic oxide as it contains O²⁻ which has high tendency to bind with hydrogen atoms. This reaction is an exothermic reaction and is conducted in cold water to produce NaOH.

The correct formula for sodium hydroxide in the given word equation is NaOH.

The word equation provided is:

[tex]\[ \text{sodium oxide} + \text{water} \rightarrow \text{sodium hydroxide} \][/tex]

To represent this equation with chemical formulas, we need to know the formulas for each compound:

- Sodium oxide (Na2O) is an ionic compound composed of sodium ions (Na+) and oxide ions (O2-).

- Water (H2O) is a molecular compound composed of two hydrogen atoms and one oxygen atom.

- Sodium hydroxide (NaOH) is an ionic compound composed of sodium ions (Na+) and hydroxide ions (OH-).

The balanced chemical equation for the reaction is:

[tex]\[ \text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} \][/tex]

This equation shows that sodium oxide reacts with water to form sodium hydroxide. The formula for sodium hydroxide is NaOH, which consists of one sodium ion (Na+) and one hydroxide ion (OH-). The coefficient 2 in front of NaOH indicates that two moles of sodium hydroxide are produced for every mole of sodium oxide and water that react.

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