Answer:
68.97g of milk must be added
Explanation:
To solve this problem we need to use the equation:
Q = C*m*ΔT -Coffee cup calorimeter equation-
This equation relates mass and change in temperature with heat of solution, Q.
The energy that decreases in the milk is the same that increases in the milk. That is:
Q(milk) = Q(tea)
Using the equation:
4.184J/molK* Mass milk * (65.0°C - 7.00°C) = 4.184J /molK * 200g (Tea) * (85.0°C - 65.0°C)
Mass milk * 58.0°C = 200g * 20.0°C
Mass milk = 4000g°C / 58.0°C
Mass milk =
68.97g of milk must be added
Look at pictures and help please
Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.
Explanation:
More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.
Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.
Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.
identify the organ system pictured below and state two functions of this system in the body
Answer:
skeletal system
Explanation:
to create and fliter blood and provide frame-work to the human body and support
Calculate the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the following balanced chemical equation: 2 Al + 6 HCl → 2 AlCl3 + 3 H2
Answer: The mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
Explanation:
The given balanced reaction equation is as follows.
[tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex]
Here, the mole ration of Al and hydrogen produced is 2 : 3
As mass of aluminum is given as 26.98 g. So, moles of aluminum (molar mass = 26.98 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{26.98 g}{26.98 g/mol}\\= 1 mol[/tex]
So, when 1 mole of Al reacted then 1.5 moles of hydrogen is produced as per the given mole ratio.
Therefore, mass of hydrogen formed is calculated as follows.
[tex]mass = moles \times molar mass\\= 1.5 mol \times 2.02 g/mol\\= 3.03 g[/tex]
Thus, we can conclude that the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
Why can light be treated like a particle?
Answer:
Light can be treated like particles because it is made of chunks like things called protons.
How much energy does an X-ray with an 8 nm (8 x 10-9m) wavelength have?
A. 1.99 x 10-25 J
B. 3.33 x 1016 J
C. 2.48 x 10-17 j
D. 8.28 x 10-26 J
Answer:
it would be option C
Explanation:
Speed of light = 3×10^8m/s
Planck's constant = 6.626×10^-34 Js
Wavelength = 8 x 10^-9 m
Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9
Energy = [19.878×10^(8-34)] / 8 x 10^-9
Energy = 2.48475 × 10^(-26+9)
Energy = 2.48×10^-17 J
What volume in mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr?
Answer: A volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.
Explanation:
Given: Moles = 0.0130 mol
Molarity = 0.220 M
Molarity is the number of moles of solute present in liter of a solution.
[tex]Molarity = \frac{moles}{volume (in L)}[/tex]
Substitute the values into above formula as follows.
[tex]Molarity = \frac{moles}{volume (in L)}\\0.220 M = \frac{0.0130 mol}{Volume (in L)}\\Volume (in L) = 0.059 L[/tex]
As 1 L = 1000 mL
So, 0.059 L = 59 mL
Thus, we can conclude that a volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.
Emily spills concentrated sodium hydroxide solution on her lab bench. What she should do first?
Answer:
Explanation: hell noo
A 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?
Answer:
Mass of sample in mg = 15,285 mg
Explanation:
Given:
Volume of urine sample = 15 ml
Density of sample = 1.019 g/ml
FInd:
Mass of sample in mg
Computation:
Mass = density x volume
Mass of sample in mg = Volume of urine sample x Density of sample
Mass of sample in mg = 1.019 x 15
Mass of sample in mg = 15.285 gram
Mass of sample in mg = 15.285 x 1,000
Mass of sample in mg = 15,285 mg
1. Complete the following chart for the following atoms:
Element
Mass
number
Number of
electrons
Number of
neutrons
Atomic
Number
(number of
protons)
Potassium
16
56
26
What is the atomic symbol (the nuclide) for the isotope with 15 protons and 16
neutrons? Your answer must be in the form shown:
AX
You must show the correct numbers for A and Z
and have the correct symbol (Z). (3 points)
Explanation:
If there are 15 protons, 15 nuclear particles of unit positive charge, then
Z
=
15
. Now
Z
≡
the atomic number
, and you look at your copy of the Periodic Table, and you find that for
Z
=
15
, the element phosphorus is specified.
But we are not finished. Along with the 15 defining protons, there are also 16 neutrally charged, massive nuclear particles, 16 neutrons, and the protons and neutrons together determine the atomic mass. The isotope is thus
31
P
, which is almost 100% abundant, and an important nucleus for
NMR spectroscopy.
How many grams of Al2O3 is extracted from 250. g of FeO?
Answer:
[tex]m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Explanation:
Hello there!
In this case, if we consider the following chemical reaction, whereby Al2O3 is produced from Al and FeO:
[tex]3FeO+2Al\rightarrow 3Fe+Al_2O_3[/tex]
Thus, since there is 3:1 mole ratio of FeO to Al2O3, it turns out feasible for us to use their molar masses, 71.844 g/mol and 101.96 g/mol respectively, to obtain the grams of the latter as follows:
[tex]m_{Al_2O_3}=250.gFeO*\frac{1molFeO}{71.844gFeO}*\frac{1molAl_2O_3}{3molFeO} *\frac{101.96gAl_2O_3}{1molAl_2O_3}\\\\m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Regards!
State three natural conditions under which plasmolysis could occur
Answer:
When the cell is placed in salt solutionWhen leave celss dryWhen cells are placed in sugar solutionExplanation:
Hope this helps
2KClO3 (s)⇄2KCl (s)+ 3O2 (aq) equilibrium constant
Answer: The equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
Explanation:
The equilibrium constant is defined as the ratio of the concentration of the products to the concentration of reactants each raised to the power of their stoichiometric coefficients.
The concentration of all the solids and liquids are considered to be 1 in the expression of equilibrium constant
For the given chemical equation:
[tex]2KClO_3(s)\rightleftharpoons 2KCl(s)+3O_2(aq)[/tex]
The expression of equilibrium constant follows:
[tex]K_{eq}=[O_2]^3[/tex]
Hence, the equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
Molecule contains carbon, hydrogen and sulfur atoms. When a sample of 0.535g of this compound is burnt in oxygen, 1.119 g of CO2and 0.229 gof H2O and 0.407g of SO2are obtained.
Calculate its empirical formula.
Answer:
The empirical formula is, C4H4S
Explanation:
Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles
Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g
Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles
Mass of hydrogen = 0.025 moles × 1 = 0.025 g
Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles
Mass of sulphur= 0.0064 moles ×32 = 0.2 g
Now we obtain the mole ratios by dividing through by the lowest ratio.
C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles
C4H4S
1. Determine the volume of SO2 (at STP) formed from the reaction of 96.7 mol FeS2 and 55.0 L of O2 at 358 K and 1.20 atm.
4 FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)
Answer:
40.0L of SO2 are produced
Explanation:
To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:
Moles O2:
n = PV/RT
n = 1.20atm*55.0L / 0.082atmL/molK*358K
n = 2.25 moles of O2.
Clearly, limiting reactant is O2.
The moles of SO2 produced are:
2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2
Volume SO2:
V = nRT/P
V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm
V = 40.0L of SO2 are produced
Según la cinética química para que una reacción ocurra, los átomos o moléculas deben
I. Chocar con la suficiente energía. II. Chocar con una concentración adecuada. III. Ser choques efectivos
A) Solo I.
B) Solo I y II.
C) Solo I y III.
D) Solo II y III.
E) I, II, III.
Answer:
solo I
Explanation:
Según esta teoría para que se produzca una reacción deben cumplirse tres condiciones: Las moléculas de los reactivos tienen que chocar entre sí. Estos choques deben de producirse con energía suficiente de forma que se puedan romper y formar enlaces químicos.
Which commercial technology commonly uses plasmas?
a radio
a race car
a television
a microwave oven
Answer:
A television is commercial technology commonly uses plasmas.A diver exhales a bubble with volume of 250 mL at pressure of 2.4 atm and temperature of 15 C. How many gas particulate in this bubble?
Answer:
1.5x10²² particulates
Explanation:
Assuming ideal behaviour, we can solve this problem by using the PV=nRT formula, where:
P = 2.4 atmV = 250 mL ⇒ 250 / 1000 = 0.250 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 15 °C ⇒ 15 + 273 = 288 KWe input the given data:
2.4 atm * 0.250 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 288 KAnd solve for n:
n = 0.025 molFinally we calculate how many particulates are there in 0.025 moles, using Avogadro's number:
0.025 mol * 6.023x10²³ particulates/mol = 1.5x10²² particulatesis it a physical or chemical change when a candle is lit
An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body weight.
How many milliliters of this suspension should be given to an infant weighing 13 lb.
Answer:
0.8853 mL
Explanation:
First we convert 13 lb to kg, keeping in mind that 1 lb = 0.454 kg:
13 lb * [tex]\frac{0.454kg}{1lb}[/tex] = 5.902 kgThen we calculate how many mg of acetaminophen should be given, using the recommended dose and infant mass:
15 mg/kg * 5.902 kg = 88.53 mgFinally we calculate the required mL of suspension, using its concentration:
88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mLYou are an intermediate product of an industrial process which intends to separate iron from its ore. A well known iron ore is hematite. Which of these ores does not contain iron?
Goethite
Malachite
Siderite
Limonite
Answer:
Malachite
Explanation:
Malachite is a copper carbonate hydroxide mineral, with the equation Cu2CO3(OH)2. This dark, green-joined mineral solidifies in the monoclinic precious stone framework, and frequently shapes botryoidal, sinewy, or stalagmitic masses, in cracks and profound, underground spaces, where the water table and aqueous liquids give the way to synthetic precipitation. So, the answer is malachite. Best of Luck!
Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide (NaN3) to decompose explosively according to the following reaction. 2 NaN3(s) --> 2 Na(s) 3 N2(g) What mass in grams of NaN3(s) must be reacted in order to inflate an air bag to 79.5 L at STP
Answer:
154 g
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP
At STP, 1 mole of N₂ occupies 22.4 L.
79.5 L × 1 mol/22.4 L = 3.55 mol
Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂
The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.
Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃
The molar mass of NaN₃ is 65.01 g/mol.
2.37 mol × 65.01 g/mol = 154 g
Why is iodine always Used in a solution containing excess I2
Answer:
If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to reduce a chemical species while simultaneously forming iodine.
Iodine always used in a solution excess KI is given to aid in the solubilization of free iodine, which would be insoluble in clean water during normal circumstances.
What is Iodine?
Iodine is a kind of element which are mainly used in iodometry titration. It can be represented by I.
What is solution?A solution would be a homogenous mixture of two components, usually a solute as well as a solvent.
Iodimetry would be a technique that uses a standard iodine solution as a titrant for such an oxidizable analyte. When an excessive amount of iodide is used to decrease a chemical while somehow producing iodine.
To know more about iodine and solution
https://brainly.com/question/16867213
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Guys I don't know science, if you are intelligent tell me what is science
Explanation:
Science is the pursuit and application of knowledge and understanding of the natural and social world following a systematic methodology based on evidence
What is the molar mass of Na2SO4?
O A. 142.04 g/mol
O B. 71.05 g/mol
O c. 238.22 g/mol
O D. 94.04 g/mol
Answer: A. 142.04 g/mol
Explanation:
Use the following information to calculate the concentration, Ka and pka for an unknown monoprotic weak acid. (8 pts.) 20.00 mL
Volume of unknown weak acid used : 20.00 mL Total volume of 0.20 M NaOH required to reach the equivalence point: 18.50 mL Initial pH of the weak acid 2.87
Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
Where 1 mole of HX reacts with 1 mole of NaOH
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
Where [H⁺] = [X⁻] because comes from the same equilibrium
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
pKa = 5.01Which of the following mixtures is best separated by the use of a separating funnel?
methane and water
ethyl ethanoate and water
ethanol and water
ethanoic acid and water
Answer:
ethyl ethanoate and water
Explanation:
At the point when one fluid doesn't blend in with another yet glides on top of it, an isolating pipe can be utilized to isolate the two fluids. Oil glides on water. This combination can be isolated utilizing an isolating channel as demonstrated on the following page.
Ethyl liquor and water are two miscible fluids. Refining is a cycle that can be utilized to isolate an unadulterated fluid from a combination of fluids. An isolating channel can be utilized to isolate the parts of the combination of immiscible fluids.
The answer is ethyl ethanoate and water. Hope this helps you!
What is its density in kilograms per cubic meter?
Copper has a density of 8950 kg/m3 = 8.95 kg/dm3 = 8.95 g/cm3. Water has a density of 1000 kg/m3 = 1000 g/L = 1 kg/dm3 = 1 kg/L = 1 g/cm3 = 1 g/mL.
HOPE IT HELPS❤️
Can someone please help with these 2?
Equilibrium shifts to the right.
OPTION A
A transition metal in the fourth period from the following list : Cu, O , Pr, Ag
Answer:
Cu
Explanation:
Groups 3 - 12 (or groups IIA - IIB) of the periodic table contain transition elements. Transaction elements start from period four (4) of the periodic table. The phrase alludes to the fact that the d sublevel is filling at a lower main energy level than the s sublevel that came before it.
The transition elements' arrangement is inverted from the fill order, with the 4 s filled prior to the actual 3 d begins. The transition elements are commonly referred to as transition metals since they are all metals. They are less reactive than the metals in Groups 1 and 2 and have normal metallic characteristics.
From the options given Cu is the only transition metal in the fourth period on the periodic table.
Lee y analiza a detalle el Anexo #1 de este plan de trabajo el cual habla sobre las fuentes alternativas de energía. Posteriormente con la información elabora un cartel o un cuadro sinóptico en tu cuaderno donde organices la información para darla a conocer a los miembros de tu comunidad.
Answer: el texto no es tan claro
