Escriba un poema usando 4 de las siguientes palabras: lluvia, nieve, lluvia, granizo, tornado, huracanes, frente cálido, frente frío, alta presión, baja presión, termómetro, barómetro, anemómetro, pluviómetro, veleta o higrómetro.
plis ayúdenme porfa
Answer:
i dont speak mexican
Explanation:
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A student performs a titration of 51.0 mL of a phosphoric
acid (H PO) solution of unknown concentration with a
standardized 1.25 M NaOH solution. The titration requires
26.2 mL of base to reach the third equivalence point. What is
the concentration of the H3PO4
solution?
From the information available in the question, the concentration of the acid is 0.21 M.
H3PO4(aq) + 3NaOH(aq) -----> Na3PO4(aq) + 3H2O(l)
Volume of acid(VA) = 51.0 mL
Concentration of acid (CA) = ?
Volume of base (VB) = 26.2 mL
Concentration of base (CB) = 1.25 M
Number of moles of acid (NA) = 1
Number of moles of base (NB) = 3
Using the formula;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 1.25 M × 26.2 mL × 1/51.0 mL × 3
CA = 0.21 M
The concentration of the acid is 0.21 M.
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A mixture is made of 40 ml of salt water to 200 ml of solution. What percent of the solution is salt water?
Answer:
16.7%
Explanation:
40 ml of salt water + 200 ml of solution = 240 ml
40/240 = 4/24 = 1/6=16.7%
A chemical reaction involves the reactants A and B, and the product C.
A+B→C
Trial 1 2 3
[A](mol/L) 0.10 0.10 0.20
[B](mol/L) 0.10 0.20 0.10
−Δ[A]Δt(molL s) 3.08×10−9 2.464×10−8 1.232×10−8
Explanation:
a+b =ab
answer
ab added all
is give answer
Which of the following best describes the scientific exploration of the atom?
. Nhiệt độ ban đầu của 344 g một mẫu sắt là 18,2oC. Nếu mẫu sắt này hấp thụ 2,25 kJ nhiệt lượng thì nhiệt độ cuối của mẫu sắt này là bao nhiêu? (CFe = 0,449 J/g.K).
Answer:
i can't understand the question
________ are neutral particles found in the nucleus of an atom.
A mixture of 3.25 moles of oxygen gas and 2.75 moles of nitrogen gas exert a total pressure of 22.4 atm. What is the partial pressure of oxygen
The partial pressure of oxygen is equal to 10.2 atm.
Given the following data:
Number of moles of oxygen = 3.25 moles.Number of moles of nitrogen = 2.75 moles.Total pressure = 22.4 atm.To determine the partial pressure of oxygen:
First of all, we would find the total number of moles of the elements:
[tex]n=3.25+2.75[/tex]
n = 6 moles
Next, we would determine the mole fraction of the oxygen by using this formula:
[tex]Molefraction \;of \;a \;substance =\frac{No.\; of \; moles \;of \;substance}{Total \;no. \;of \; moles \;of \;substances}[/tex]
Substituting the values, we have:
[tex]Molefraction \;of \;a \;substance =\frac{2.75}{6} \\\\Molefraction \;of \;a \;substance =0.4583[/tex]
For oxygen:
[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.4583 \times 22.4[/tex]
Partial pressure of oxygen = 10.2 atm.
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Answer:
12.1
Explanation:
3.25 + 2.75= 6 moles total
then take: 3.25/6 to get 0.542
multiply 0.542 by 22.4 to get 12.1
the other answer solved for nitrogen, not oxygen.
If 5 g of sodium chloride saturates 12.5 g of water at 10 °C, what mass of sodium chloride would saturate 50 g of water at constant temperature?
Explanation:
since 5g saturates 12.5g of water at 10°c
so......x would saturate 50g of water at 0°c
then you can cross multiply
Determine the mass in grams of 5.15 × 10²¹ atoms of chromium. (The mass of one mole of chromium is 52.00 g.)
The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
From Avogadro's hypothesis,
6.02×10²³ atoms = 1 mole of Cr
But:
1 mole of Cr = 52 g
Thus, we can say that:
6.02×10²³ atoms = 52 g of Cr
With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:
6.02×10²³ atoms = 52 g of Cr
Therefore,
5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³
5.15×10²¹ atoms = 0.44 g of Cr
Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
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The box shows the chemical formula for bleach.
a material that is not a mixture; has the same properties all the way through
Answer:
Explanation:
The material that is not a mixture; it has the same properties all the way through is called a substance. Thus the material that is not a mixture; it has the same properties all the way through is called a substance.
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Compare and contrast model A with model C. How are they alike: How are they different?
A) They are composed of different elements in the same ratio.
B) They are composed of the same elements in different ratios.
C) They are composed of different elements in different ratios
D) The are composed of the same elements but the bond types differ
Answer:
Model C has more double the oxygen than A. I think it is B
Explanation:
The sum of the number of proteins and neutrons in an atoms nucleus is its __________ ___________.
Answer:
Mass Number
Explanation:
In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.
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Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.
This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.
Firstly, we can write out the chemical equation to be considered:
[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]
Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:
[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]
Afterwards, we set up an equilibrium expression for this chemical reaction:
[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]
Which can be written in terms of the reaction extent, [tex]x[/tex]:
[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]
Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:
[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]
Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.
Thus, the required concentrations at equilibrium are about:
[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]
[tex][Ni^{2+}]=0M[/tex]
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https://brainly.com/question/13043707Grace wanted to find out the best conditions for growing lettuce plants.
She took 4 trays and planted 8 lettuce plants in each.
The results of her investigation are shown below.
How many days did the investigation last?
Use the table to help you.
Explanation:
the investigation lasts for 7 days !
hope this helps you.
The anion O3− does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Indicate the values of nonzero formal charges and include lone pair electrons.
One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.
Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.
We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.
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2. Alex pulls on the handle of a claw hammer with a force of 15 N. If
the hammer has a mechanical advantage of 5.2, how much force
is exerted on the nail in the claw?
Answer:
78n
Explanation:
The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.
What is the mechanical advantage?The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.
The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.
Given, the input force = 15 N
The mechanical advantage of the hammer = 5.2
Mechanical advantage = Output force/ Input force
5.2 = Output/15
Output force = 15 ×5.2 = 78 N
Therefore, the force is exerted on the nail in the claw is equal to 78 N.
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pls help 8th grade k12
A metal (FW 341.1 g/mol) crystallizes into a body-centered cubic unit cell and has a radius of 1.74 Angstrom. What is the density of this metal in g/cm3
This problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.
Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:
[tex]\rho =\frac{Z*M}{V*N_A}[/tex]
Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:
[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]
And finally the required density in g/cm³:
[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]
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Each element can be indentified by the number of _______ found in its nucleus, which also equals the elements _______ _______.
6) 0.5 moles of gas is kept at 2.0 L of volume and 0.75 atm of pressure. What is the temperature of the gas in K?
Answer:
310K
Explanation:
Rearrange PV=nRT to get T=PV/nR
T=(2.0L)(0.75atm)/(0.5mol)(0.08206)
=36.5 or 37
add 273 for K to get
310
GIVING BRAINLY AND 20 POINTS
A sound wave in air has the wavelength of 1.36 m. Calculate its frequency? Assume the speed is 340 m/s.
Answer: the answer is 0.074Hz
Explanation:
Given, (In air)
Velocity V=340m/s
Frequency f=20,000Hz
Wavelegth λ=?
V=f.λ
λ=
F
V
=
20,000
340
=0.017Hz
Also, Given (in Water)
Velocity, V=1480m/s
Frequencyf=20,000Hz
wavelength, λ=?
V=F.λ
λ=
F
V
=
20,000
1480
=0.074Hz
Answer:
frequency
Explanation:
frequency is velocity/ wavelength
340/1.36
250
what is the expected hybridization of the central atom tetrahedral
Answer:
Tetrahedral molecules are normally spy hybridized.
Explanation:
Compound A is neutral and Compound B is acidic. Both are water-insoluble solids. A and B are dissolved in dichloromethane (DCM) and extracted with aqueous base. The layers are then separated. What must be done to obtain the compound in the aqueous layer
In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.
When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.
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in a certain reaction, Fe and O2 combine to form iron (iii) oxide. 14.7 moles of Fe and 13.0 moles of O2 are placed in a container and the reaction proceeds iwth 100% yield. which is the excess reactant
Answer:
Fe is the excess reactant
Reaction yields are the amount of the reactant and the products of a chemical reaction. In the reaction between iron and oxygen, iron is the excess agent.
What is excess reactant?The reactant present in an extra quantity than the other reactant in a chemical reaction which reacts with the limiting reactant is called an excess reactant.
In a reaction mixture, the excess reactant is present even when the limiting agent is completely consumed.
The chemical reaction between iron and oxygen is shown as,
[tex]\rm 4 Fe(s) + 3 O_{2}(g) \rightarrow 2 Fe_{2}O_{3}[/tex]
From the reaction, it can be said that oxygen is a limiting reagent that limits the formation of iron (iii) oxide.
Therefore, iron is an excess reactant.
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Is brass a solution:
Answer:
Brass is an alloy, and either a "solid solution".
Alloys in general may be solid solutions or they simply be mixtures
Explanation:
Hope it Helps you!!How many grams of sodium acetate NaCH3COO, , must be dissolved to prepare 200. mL of a 0.216 M aqueous solution of the salt? g
Answer:
Explanation:
You have to find the mass in the mole before the molarity and then in the mole.
) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at 25 °C, and the density of water at 25 °C is 0.9970 g>mL)
At the left, grains of the mineral calcite float on the surface of the liquid bromoform (d = 2.890 g/mL) At the right, the grains sink to the bottom of liquid Chloroform (d = 1.444 g/mL). By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined
Hey there!
It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.
In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:
[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]
Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:
[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]
Thus, the density of the calcite sample will be:
[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]
This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL
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