The identity of the sample is copper. The correct option is D. copper
To determine the identity of the sample, we will determine its specific heat, c.
From the formula
Q = mcΔT
Where Q is the quantity of heat
m is the mass
c is the specific heat of the substance
and ΔT is the change in temperature
From the given information
Q = 3080 J
m = 200 g
ΔT = T₂ - T₁ = 60°C - 20°C = 40°C
Putting the parameters into the formula, we get
3080 = 200 × c × 40
3080 = 8000c
∴ c = 3080 ÷ 8000
c = 0.385 J / g°C
From specific heat table, the substance that has a specific heat of 0.385 J / g°C is Copper.
Hence, the identity of the sample is copper. The correct option is D. copper
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Many enjoys the warm waters of a ______.
Geyser
Volcano
Hot spring
Furmarole
Tectonic
Answer:
hot spring
Explanation:
The mineral water in hot springs can also help reduce stress by relaxing tense muscles. Meanwhile as your body temperature rises in the bath, and then cools once you exit can also help you relax and fall into a deeper sleep.
What is the Ksp expression for the dissociation of calcium oxalate?Immersive Reader
(4 Points)
Ksp=[Ca⁺²] x [C₂O₄⁻²]
Ksp=[Ca⁺²]² x [C₂O₄⁻²]
Ksp=[Ca⁺²]⁴ x [C₂O₄⁻²]
Ksp=[Ca⁺²] x [C₂O₄⁻²]²
Answer:
Ksp = [Ca⁺²] × [C₂O₄⁻²]
Explanation:
Step 1: Write the balanced reaction for the dissociation of calcium oxalate
CaC₂O₄(s) ⇄ Ca⁺²(aq) + C₂O₄⁻²(aq)
Step 2: Write the expression for the solubility product constant (Ksp) of calcium oxalate
The solubility product constant is the equilibrium constant for the dissociation reaction, that is, it is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It doesn't include solids nor pure liquids because their activities are 1.
Ksp = [Ca⁺²] × [C₂O₄⁻²]
2 Al(s) + Fe2O3(aq) - AlO3(aq) + 2 Fe(s)
You react 20.00 grams of aluminum with iron(III) oxide. How many grams of iron should you produce?
What is the percent yield if the experimental yield is 32.67 grams of iron?
Answer:
78.8%
Explanation:
The equation of the reaction is;
2 Al(s) + Fe2O3(aq) ------> Al2O3(aq) + 2Fe(s)
Number of moles in 20g of Al= 20g/27 g/mol = 0.74 moles
From the reaction equation;
2 moles of Al yields 2 moles of Fe
0.74 moles of Al yields 0.74 moles of Fe
Hence;
Mass of Fe produced = 0.74 moles of Fe * 56 g/mol
Mass of Fe produced = 41.44 g of Fe (This is the theoretical yield of Fe)
percent yield = actual yield/ theoretical yield * 100
actual yield = 32.67 grams of iron
percent yield = 32.67 g/41.44 g * 100
percent yield = 78.8%
The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?
Answer: The concentration of hydrogen ions for this solution is [tex]1.99 \times 10^{-3}[/tex].
Explanation:
Given: pOH = 11.30
The relation between pH and pOH is as follows.
pH + pOH = 14
pH + 11.30 = 14
pH = 14 - 11.30
= 2.7
Also, pH is the negative logarithm of concentration of hydrogen ions.
[tex]pH = - log [H^{+}][/tex]
Substitute the values into above formula as follows.
[tex]pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}[/tex]
Thus, we can conclude that the concentration of hydrogen ions for this solution is [tex]1.99 \times 10^{-3}[/tex].
In an experiment, a student wants to increase the rate of a reaction that involves gases. Which change to the reactants would accomplish this? Increase the volume to decrease pressure and to increase concentration. Increase the volume to increase pressure and to decrease concentration. Decrease the volume to decrease pressure and to increase concentration. Decrease the volume to increase pressure and to increase concentratio
Answer:
Decrease the volume to increase pressure and to increase concentration
Explanation:
A gaseous reaction is affected by the volume of the reactants. To increase the rate of the forward reaction, we need to ensure that the volume of the reactants is manipulated in such a way that the forward reaction is favored.
Thus, when the volume of reactants is decreased, the pressure increases.This increase in pressure has the same effect as increase in concentration. Hence, the rate of forward reaction increases.
what causes salinization
Answer:
Salinization is the increase of salt concentration in soil and is, in most cases, caused by dissolved salts in the water supply. This supply of water can be caused by flooding of the land by seawater, seepage of seawater or brackish groundwater through the soil from below.
Explanation:
The system we will be observing in this lab involves the cobalt(II) ion. This
ion can form two complexes, one with water and one with the chloride
ion (CI'). Below is a reaction showing the addition of chloride ions to the
cobalt - water complex.
Co(H2016 + Cl <--> COC142- + H2O
2+
pink
blue
What happened when HCl was added?
Answer:
See Explanation
Explanation:
In the equilibrium shown in the question,on the left hand side is a pink hexaaquacobalt II complex which is pink in colour. On the right hand side, we have a tetrachlorocobalt II complex which is blue in colour.
The equilibrium can be shifted from left to right by adding HCl (a source of chloride ions).
Hence, when HCl is added to the system, the equilibrium position is shifted to the right thereby yielding the blue solution of the tetrachlorocobalt II complex.
What is the balanced chemical equation for the reaction between aluminum and copper nitrate? Use this resource on polyatomic ions and the periodic table to help you.
a. An aqueous solution of Mn(NO3)2 is very pale pink, but an aqueous solution of K4[Mn(CN)6] is deep blue. Explain why the two differ so much in the intensities of their colors.
b. Predict which of the following compounds would be colorless in aqueous solution:
a. K2[Co(NCS)4]
b. Zn(NO3)2
c. [Cu(NH3)4]Cl2
d. CdSO4
e. AgClO3
f. Cr(NO3)2
Answer:
See Explanation
Explanation:
The colour of many transition metal complexes stem from transitions of electrons between energy levels. These transitions are governed by the spin selection rules and the colour is determined by the magnitude of crystal field splitting.
According to the spin selection rules, transitions in which ΔS = 0 are forbidden. Hence, a Mn^2+high spin compound is expected to be colourless. However, contrary to the spin selection rules Mn^2+high spin compounds do exhibit transitions in which the intensity is only about one-hundredth of the intensity of the spin allowed transitions. Thus many Mn^2+ high spin compounds such as Mn(NO3)2 are very pale pink or off white.
Note also that the crystal field stabilization energy of Mn^2+ which is a d^5 low spin ion is zero hence the very pale colour observed.
K4[Mn(CN)6] is deep blue as a result of charge transfer. Also, the compound exhibits an observed crystal field stabilization energy because it is a d^5 low spin compound hence the observed colour. Its low spin nature is because the cyanide ion is a strong field ligand hence it causes a greater magnitude of crystal filed splitting.
The following compounds are colourless;
Zn(NO3)2
CdSO4
AgClO3
One thing that is common to all the compounds listed above is that they are all d^10 compounds. This means that they all possess completely filled d-orbitals hence they are colourless.
1. Calculate and interpret the equilibrium constant. Using the reaction below.
The equilibrium concentrations 0.60 M for E, 0.80 M for F, and 1.30 M for G. (Note: E, F, and G are all gases.) Do not include your solution.
Answer:
kc = [G]² / [E] [F], kc = [1.30M]² / [0.60M] [0.80M]
Explanation:
The reaction is:
E + F ⇄ 2G
The equilibrium constant, kc, must be written as the ratio of the molar concentrations of products over reactants. Each concentration powered to its coefficient.
For the reaction of the problem, kc is:
kc = [G]² / [E] [F]Replacing the given concentrations:
kc = [1.30M]² / [0.60M] [0.80M]Please help
Apply your knowledge and understanding of equilibrium constant in solving the following problems:
The equilibrium constant Kc for the reaction below is 170 at 500 K.
Determine whether the reaction mixture is at equilibrium when the concentrations of the components at this temperature are as follows:
[N2]=1.50
[H2]=1.00
[NH3]=8.00
If it is not at equilibrium, state and explain in which direction the reaction will proceed.Multi Line Text.
2()+32() ⇄ 23()
Answer:
The reaction will proceed to the right to attain the equilibrium.
Explanation:
Step 1: Write the balanced equation
2 N₂(g) + 3 H₂(g) = 2 NH₃(g)
Step 2: Calculate the reaction quotient
The reaction quotient (Qc) is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.
Qc = [NH₃]² / [N₂]² × [H₂]³
Qc = 8.00² / 1.50² × 1.00³ = 28.4
Since Qc ≠ Kc, the reaction is not at equilibrium.
Since Qc < Kc, the reaction will proceed to the right to attain the equilibrium.
The spectrochemical series is I < Br< < Cl^- < F
The complex [Ni(C1)6.]^4- is green and the complex [Ni(en)3]2^+ is violet, where en corresponds to the ethylenediamine ligand. The reason for the difference in the color of these complexes is:
a. The [Ni(Cl)6]4- absorbs green light, and the [Ni(en)3]2+ absorbs violet light.
b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
c. The chloride ligand is green, and the ethylenediamine ligand is violet.
d. The splitting of the d-orbitals is larger in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
e. Two of the above statements are correct.
f. The difference in the coordination numbers of the chloride complex and the en complex.
g. The difference in the oxidation states of the central transition metals in the [Ni(Cl)6]4- complex and the [Ni(en)3]2+ complex.
Answer:
b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
Explanation:
The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.
Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.
Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.
Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
Estimate the volume of a solution of 5M NaOH that must be added to adjust the pH from 4 to 9 in 100 mL of a 100 mM solution of a phosphoric acid?
Answer:
3mL of 5M NaOH must be added to adjust the pH to 7.20
Explanation:
When NaOH is added to phosphoric acid, H₃PO₄, the reaction that occurs are:
NaOH + H₃PO₄ ⇄ NaH₂PO₄ + H₂O pKa1 = 2.15
NaOH + NaH₂PO₄ ⇄ Na₂HPO₄ + H₂O pKa2 = 7.20
NaOH + Na₂HPO₄ ⇄ Na₃PO₄ + H₂O pKa3 = 12.38
We can adjust the pH at 7.20 = pKa2 if NaH₂PO₄ = Na₂HPO₄. To make that, we must convert, as first, all H₃PO₄ to NaH₂PO₄ and the half of NaH₂PO₄ to Na₂HPO₄. To solve this question we need to find the moles of phophoric acid in the initial solution. 1.5 times these moles are the moles of NaOH that must be added to fix the pH to 7.20:
Moles H₃PO₄:
100mL = 0.100L * (0.100mol / L) = 0.0100 moles H₃PO₄
Moles NaOH:
0.0100 moles H₃PO₄ * 1.5 = 0.0150 moles NaOH
Volume NaOH:
0.0150 moles NaOH * (1L / 5moles) = 3x10⁻³L 5M NaOH are required =
3mL of 5M NaOH must be added to adjust the pH to 7.203 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
What is pH?It is the negative log of the concentration of Hydrogen ions in the solution.
To calculate the volume of NaOH first, calculate the moles of NaOH and H₃PO₄.
Moles of H₃PO₄.
[tex]\rm moles \ of \ H_3PO_4 = 100\rm \ mL = 0.100\rm \ L \times (0.100 \ mol / L)\\\\\rm moles \ of \ H_3PO_4 = 0.01[/tex]
The moles of NaOH:
[tex]\rm Moles \ of \ NaOH =0.01 \ moles \ H_3PO_4\times 1.5 \\\\\rm Moles \ of \ NaOH= 0.0150[/tex]
The volume of NaOH:
[tex]\rm Volume\ of \ NaOH = \rm 0.0150\ moles\ NaOH \times (1 \ L / 5 \ moles) \\\\\rm Volume\ of \ NaOH = 3\times 10^{-3} L[/tex]
Therefore, 3 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
Learn more about pH:
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Thermal energy naturally flows from _________ matter to _______ matter.
Answer:
Warmer
Cooler
Explanation:
WO=C=Ộ
What elements are in this compound
Answer:
Oxygen Carbon DioxideExplanation:
The compound shown is Carbon Dioxide and is as a result of a covalent bond between Carbon and Oxygen where carbon is sharing two electrons each with the oxygen atoms.
The O is the atomic symbol for Oxygen and the C is the symbol for Carbon. Carbon Dioxide then has the atomic symbol of CO₂ to show the relative number of atoms of each element in it.
What is the function of a
catalyst?
A. Build enzymes
B. Speed up chemical reactions
C. Regulate the function of an enzyme
I believe it is to speed up chemical reactions
define saturated and unsaturated fats
Answer:
Saturated and unsaturated:-Are a form of fat in which all or most of the fatty acid chains are single bonds. Glycerol and fatty acids are the two types of smaller molecules that make up fat.
Saturated fat is found in:cakessausagescheesebutterExamples of unsaturated fats:- OliveNuts(almonds, hazelnuts)Seeds(pumpkin and sesame seeds)hope it helps...
URGENT
Best answer gets Brainliest!
Mg+ 2HCI ——> MgCI2+ H2
Answer:
Reaction type: Single displacement
Reactant: Magnesium
Product: Dihydrogen - H2
When 214.5 g of calcium carbonate react with 321.9 g of aluminum fluoride, how many grams of each product can be produced?
Answer:
grams CaF₂ = grams Al₂(CO₃)₃ = 167.7 grams (4 sig. figs.)
Explanation:
3CaCO₃ + 2AlF₃ => 3CaF₂ + Al₂(CO₃)₃
Given: 214.5g/(100g/mol) 321.9g/(84g/mol) =>2.15 moles & 0.71 mole
= 2.15 mole = 3.83 mole => ?g CaF₂ ?g Al₂(CO₃)₃
ID of Limiting Reactant => divide mole values by respective coefficient & smaller value is the Limiting Reactant.
CaCO₃ => (2.15/3) = 0.72 and AlF₃ => (3.83/2) = 1.92
The value for CaCO₃ (=0.72) < the value for AlF₃ (=1.92) => CaCO₃ is the Limiting Reactant
Note: When working problem using mole ratios, use mole value for Limiting Reactant and not the value used to determine Limiting Reactant, in this case 2.15 moles CaCO₃.
grams CaF₂ Produced:
moles CaF₂ produced = 3/3(2.15) moles CaF₂ = 2.15 moles CaF₂
grams CaF₂ produced = 2.15 moles CaF₂ x 78 g CaF₂ / mole CaF₂
= 167.7 grams CaF₂
grams Al₂(CO₃)₃ produced:
moles Al₂(CO₃)₃ produced = 1/3(2.15) moles Al₂(CO₃)₃ = 0.72 mole Al₂(CO₃)₃
grams Al₂(CO₃)₃ = 0.72 mole Al₂(CO₃)₃ x 234 g Al₂(CO₃)₃ / mole Al₂(CO₃)₃ = 167.7 grams Al₂(CO₃)₃
At what temperature, in Celsius, will 88.0 g of Ne exert a pressure of 350. kPa in a
48.5 L container? NOTE: You must show your calculation on the attached scratch
paper, including which of the Gas Law formulas you used. *
Answer:
191 °C
Explanation:
We'll begin by calculating the number of mole in 88.0 g of Ne. This can be obtained as follow;
Mass of Ne = 88.0 g
Molar mass of Ne = 20 g/mol
Mole of Ne =?
Mole = mass /molar mass
Mole of Ne = 88 / 20
Mole of Ne = 4.4 moles
Next, we shall determine the temperature. This can be obtained as follow:
Mole of Ne = 4.4 moles
Pressure (P) = 350 KPa
Volume (V) = 48.5 L
Gas constant (R) = 8.314 KPa.L/Kmol
Temperature (T) =?
PV = nRT
350 × 48.5 = 4.4 × 8.314 × T
16975 = 36.5816 × T
Divide both side by 36.5816
T = 16975 / 36.5816
T = 464 K
Finally, we shall convert 464 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 464 K
T(°C) = 464 – 273
T(°C) = 191 °C
Thus, the temperature is 191 °C
Calculate the energy of an electron in the n = 2 level of a hydrogen atom.
Answer: The energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
Explanation:
Given: n = 2
The relation between energy and [tex]n^{th}[/tex] orbit of an atom is as follows.
[tex]E = - \frac{13.6}{n^{2}} eV[/tex]
Substitute the values into above formula as follows.
[tex]E = - \frac{13.6}{n^{2}} eV\\= - \frac{13.6}{(2)^{2}}\\= - 3.40 eV[/tex]
The negative sign indicates that energy is being released.
Thus, we can conclude that the energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
A 0.682 g sample of a weak monoprotic acid, HA was dissolved in sufficient water to make 50.0 mL of solution and was titrated with a 0.135 molar NaOH solution. After the addition of 10.6 milliliters of base, a pH of 5.65 was recorded. The equivalence point was reached after the addition of 27.4 milliliters of the 0.135 molar NaOH.
a. Calculate the number of moles of acid in the original sample.
b. Calculate the molar mass of the acid HA.
c. Calculate the [H3O+] at pH = 5.65
d. Calculate the number of moles of unreacted HA remaining in solution when the pH was 5.65.
e. Calculate the value of the ionization constant, Ka, of the acid HA.
f. Calculate the value of the ionization constant, Kb, and explain how you would use it to determine the pH of a solution of a known mass of the sodium salt (Na)(A) dissolved in a known volume of water.
If the plant population decreased, the amount of carbon in the atmosphere would _______.
Increase
Stay the same
Decrease
Answer:
increase answer
Explanation:
i hope that is right
A balloon can be inflated with 4.23 Liters of O2gas at STP. How much will the balloon weigh in grams?
Answer: 6.04 g
Explanation: In STP conditions, amoint of substance
n = V / Vm = 4.23 l / 22.41 l/mol = 0.188755 mol
Molar mass M(O2) = 32 g/mol and
mass m = nM
Un recipiente cerrado, de 4,25 L, con tapa móvil, contiene H2S(g) a 740 Torr y 50,0°C. Se introduce en ese recipiente N2(g) a temperatura y presión constantes, de manera que el volumen final es el doble del volumen inicial. Calcular la cantidad de N2(g) en el recipiente, expresada en moles.
porfi ayuda
Answer:
[tex]n_{N_2}=6.41mol[/tex]
Explanation:
¡Hola!
En este caso, teniendo en cuenta la información dada por el problema, inferimos que primero se debe usar la ecuación del gas ideal con el fin de calcular las moles de gas que se encuentran al inicio del experimento:
[tex]PV=nRT\\\\n=\frac{RT}{PV} \\\\n=\frac{0.08206\frac{atm*L}{mol*K}*(50.0+273.15)K}{740/760atm*4.25L}\\\\n=6.41mol[/tex]
Seguidamente, usamos la ley de Avogadro para calcular las moles finales, teniendo el cuenta que el volumen final es el doble del inicial (8.50 L):
[tex]n_2=\frac{6.41mol*8.50L}{4.25L}\\\\n_2=12.82mol[/tex]
Quiere decir que las moles de N2(g) que se agregaron son:
[tex]n_{N_2}=12.81mol-6.41mol\\\\n_{N_2}=6.41mol[/tex]
¡Saludos!
Mixed Practice:
For questions 6-9, you may not have to use your entire flow chart. Decide where to start and end using your flow chart.
4NH3 + 5O2 → 4NO + 6H2O
MM NH3 = 17.04 g/mol; MM O2 = 32 g/mol; MM NO = 30.01 g/mol; MM H2O = 18.02 g/mol
In the above reaction, how many moles of NO are formed if 824 g of NH3 react?
27.5 mol
3510 mol
48.4 mol
774 mol
Given
Mass of NO - 824 g
Molar mass of NO - 30.01g/mol
No of moles of NO = Given mass/Molar mass
No of moles of NO = 824/30.01= 27.45 mole
Hence 27.5 moles of NO are formed!
7. Which of the following is true about rocks?
they are composed of only one mineral
they throw themselves
They contain no mineral matter
most rocks are a mixture of minerals
Answer:
most rocks are a mixture of minerals
What is the electronegativity difference
between sodium and chlorine?
Answer: 2.23 is the difference.
Explanation: Sodium has an electronegativity of 0.93 and Chlorine has an electronegativity of 3.16, so when Sodium and Chlorine form an ionic bond, in which the chlorine takes an electron away from sodium, forming the sodium cation, Na+, and the chloride anion, Cl-.
Which of the following has the correct name for either the acid or base
Answer:
Explanation:
harmless
...............
Which group of elements will form molecular compounds? (Choose all that apply)