Answer:
????????????????????.......??????????????
What happens to a linguist when it releases enough energy
Answer:
When a liquid releases enough energy, the liquid will freeze and turn into a solid. Phase changes are what causes one state of matter to turn into another. All phase changes are physical changes.
Explanation:
if ur really saying liquid a person accomplished in languages
especially : one who speaks several languages ?
please mark as brainliest
____ Al + ____ Cr2O3 ____ Al2O3 + ____ Cr
If the reaction of 23.3 grams of Cr2O3 produces 5.35 grams
of Al2O3, what is the percent yield?
Answer:
[tex]Y=34.3\%[/tex]
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
[tex]2Al + Cr_2O_3 \rightarrow Al_2O_3 + 2Cr[/tex]
We notice there is a 1:1 mole ratio between Al2O3 and Cr2O3; thus, the following stoichiometric setup is used to compute the theoretical yield first:
[tex]m_{Al_2O_3 }=23.3gCr_2O_3*\frac{1molCr_2O_3}{151.99gCr_2O_3} *\frac{1molAl_2O_3}{1molCr_2O_3} *\frac{101.96gAl_2O_3}{1molAl_2O_3} \\\\m_{Al_2O_3}=15.6gAl_2O_3[/tex]
Thus, the percent yield turns out:
[tex]Y=\frac{5.35g}{15.6g} *100\%\\\\Y=34.3\%[/tex]
Best regards!
What is the edge length of a 305-g iron cube? (The density of iron is 7.86 g/cm3g/cm3, and the volume of a cube is equal to the edge length cubed.
Answer: edge length = 3.61 cm
Explanation:
Givens
m = 369 gm
d = 7.86 gm / cm^3
v = ?
Formula
v = m / d
Solution
v = 369 / 7.86
v = 46.95 cm^3
============
That's not the answer. You must take the cube root of the volume
side = cube root (46.95)
side = 3.61 cm
Which of Newton's Three Laws is BEST described by the following statement: A ball dropped from the 10th story of a hotel had more force than when the same ball was dropped from the 5th story of the same hotel
Answer:
I believe the answer is Newton's Second Law
Explanation:
Newton's Second Law states that the acceleration of an object will count on how much mass and the amount of force that is applied.
The bowling ball was dropped from the 10th story window and had more force than when the ball was dropped from the 5th story window because the 10th story window gave the bowling ball more acceleration.
Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.
Answer:
(a). 46,666.7 g/mol; 78,571.4 g/mol
(b). 86950g/mol; 46,666.7 g/mol.
(c). 86950g/mol; 43,333.33 g/mol
Explanation:
So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.
Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.
The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.
(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.
Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.
(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.
The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.
How many atoms of hydrogen are present in 10g of methane?
Answer:
[tex]atoms\ H=1.50x10^{24}[/tex]
Explanation:
Hello!
In this case, since the molecular formula of methane is CH₄, can compute the moles of methane in 10 g first:
[tex]n_{CH_4}=10gCH_4*\frac{1molCH_4}{16.05gCH_4} =0.623molCH_4[/tex]
Now, since 1 mol of methane contains 4 mol of hydrogen atoms, and 1 mol of any atom contains 6.022x10²³ atoms via the Avogadro's number, we obtain the following number atoms of hydrogen:
[tex]atoms \ H=0.623molCH_4*\frac{4molH}{1molCH_4}*\frac{6.022x10^{23}atoms\ H}{1molH}\\\\atoms\ H=1.50x10^{24}[/tex]
Best regards!
Ge(NO2)2 whats the name of this formula?
Answer:
Germanium monoxide
hope this helps :)
Convert the volume, nm^3, to liters (L) by using the box dimensions of 4nm x 8.75nm x 10nm. Use the conversion factor of 1 L = 1 dm^3. Do not convert directly from nm^3 to dm^3
Answer:
[tex]V=3.50x10^{-22}L[/tex]
Explanation:
Hello there!
In this case, given the dimensions of the box, we first compute the volume by multiplying each side:
[tex]V=4nm*8.75nm*10nm=350nm^3[/tex]
Next, we apply the following conversion factor in order to obtain the corresponding liters:
[tex]V=350nm^3*(\frac{1m}{10^9nm} )^3*\frac{1000L}{1m^3} \\\\V=350nm^3*\frac{1m^3}{10^{27}nm^3} *\frac{1000L}{1m^3}\\\\V=3.50x10^{-22}L[/tex]
Best regards!
unscramble the answer
The components of a simple circuit are the following:
1. SURCOURFOECRTEN
2. ICEEVD
3. RISEW
4. CSITHW
Answer:
1counterforces
2device
3wires
4switch
Use the drop-down menus to label the graph
according to its type.
Answer:
relation (but not a function)
Explanation:
c on edge
The hydronium ion concentration of an aqueous solution of 0.502 M pyridine (a weak base with the formula C5H5N) is
Answer:
1.99×10¯¹⁴ M
Explanation:
We'll begin by writing the balanced dissociation equation of pyridine. This is illustrated below:
C₅H₅N + H₂O <=> C₅H₆N⁺ + OH¯
From the balanced equation above,
1 mole of C₅H₅N produced 1 mole of OH¯.
Therefore, 0.502 M C₅H₅N will also produce 0.502 M OH¯.
Finally, we shall determine the concentration of hydronium ion, H₃O⁺ in the solution. This can be obtained as follow:
Concentration of Hydroxide ion [OH¯] = 0.502 M
Concentration of hydronium ion [H₃O⁺] =?
[H₃O⁺] [OH¯] = 1×10¯¹⁴
[H₃O⁺] × 0.502 = 1×10¯¹⁴
Divide both side by 0.502
[H₃O⁺] = 1×10¯¹⁴ / 0.502
[H₃O⁺] = 1.99×10¯¹⁴ M
Thus, the concentration of hydronium ion, H₃O⁺ in the solution 1.99×10¯¹⁴ M
2) A student performs this reaction in the laboratory, and collects the hydrogen gas over water. The student collects 215.8 mL of gas. The total pressure is 755.2 mmHg and the temperature is 22.0 oC. c) Use the ideal gas law to calculate how many moles of hydrogen were produced in the reaction
Answer: 0.0086 moles of hydrogen are produced.
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
total pressure = 755.2 mm Hg
pressure of water at [tex]22^0C[/tex] = 19.8 mm Hg
P = pressure of hydrogen = (755.2-19.8) mm Hg = 735.4 mm Hg = 0.97 atm
(760 mm Hg = 1atm)
V = Volume of gas = 215.8 ml = 0.2158 L
n = number of moles
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]22.0^0C=(22.0+273)K=295K[/tex]
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{0.97atm\times 0.2158L}{0.0821 L atm/K mol\times 295K}=0.0086moles[/tex]
Thus 0.0086 moles of hydrogen are produced.
Electrons help create bonds between elements as they lie on the outside of an
atom.
If you are doing stoichiometry and it asks you to go from grams of one substance to grams of a new substance, how many steps (how many boxes) would you draw in your setup?
a.1
b.2
c.3
d.4
You do not have to explain if you want to.
Answer:
a.bc
d
a
d
i think it is it or east be responsible
make me the brainyst please
1.
Use the Rydberg equation to calculate the energy (in J) and then find the wavelength
in (nm) for the following transitions.
-18
AE = 2.179x10
a.
3-2.
Answer:
Approximately [tex]3.03\times 10^{-19}\; \rm J[/tex], which corresponds to a wavelength of approximately [tex]6.56 \times 10^{2}\; \rm nm[/tex] in vacuum.
Explanation:
When the electron of a hydrogen atom transits from energy level [tex]n_1[/tex] to [tex]n_2[/tex], the energy change would be:
[tex]\displaystyle \Delta E \approx 2.179 \times 10^{-18}\; {\rm J} \, \left(\frac{1}{{(n_1)}^{2}} - \frac{1}{{(n_2)}^{2}}\right)[/tex].
For the transition from [tex]n_1 =3[/tex] to [tex]n_2 = 2[/tex]:
[tex]\begin{aligned}\Delta E &\approx 2.179 \times 10^{-18}\; {\rm J} \, \left(\frac{1}{{(n_1)}^{2}} - \frac{1}{{(n_2)}^{2}}\right)\\ &= 2.179 \times 10^{-18}\; {\rm J} \, \left(\frac{1}{{3}^{2}} - \frac{1}{{2}^{2}}\right) \\ &\approx -3.02639 \times 10^{-19} \; \rm J \\ &\approx -3.03\times 10^{-19}\; \rm J\end{aligned}[/tex].
The value of [tex]\Delta E[/tex] is negative because energy is released during this transition.
Look up Planck's Constant (for finding frequency from energy) and the speed of light in vacuum (for finding wavelength from frequency.)
Planck's Constant: [tex]h \approx 6.62607 \times 10^{-34} \; \rm J \cdot s^{-1}[/tex].Speed of light in vacuum: [tex]c \approx 2.99792 \times 10^{8}\; \rm m \cdot s^{-1}[/tex].Calculate the frequency [tex]f[/tex] of photons from this transition using the Planck-Einstein relation:
[tex]E = h \cdot f[/tex].
Therefore:
[tex]\begin{aligned}f &= \frac{E}{h}\end{aligned}[/tex].
Calculate the wavelength [tex]\lambda[/tex] of these photos in vacuum:
[tex]\begin{aligned}\lambda &= \frac{c}{f} \\ &= \frac{c}{ E/ h} \\ &= \frac{h \cdot c}{E} \\ &\approx \frac{6.62607 \times 10^{-34}\; \rm J \cdot s \times 2.99792 \times 10^{8}\; \rm m \cdot s^{-1}}{3.02639 \times 10^{-19}\; \rm J} \\ &\approx 6.65 \times 10^{-7}\; \rm m \\ &= 6.65 \times 10^{-7}\; \rm m \times \frac{10^{9}\;\rm nm}{1\; \rm m} \\ &= 6.65 \times 10^{2}\; \rm nm\end{aligned}[/tex].
niling Reactants
POSSIBLE POINTS: 4.55
6 K + Cr2S3 ----> 2 Cr + 3 KZS
A scientist combines 151 grams of potassium and 151 grams of chromium III sulfide, how many moles of chromium will be
made?
(The next question will ask about the limiting and excess reactants for this reaction)
O 11.6 g
O 0.377 g
O 1.29 g
O 1.51 g
Given the following balanced chemical equation:
N2(g) + 3H2(g) + 2NH3(8)
What is the maximum amount of NH3(g) that can be produced from 2.0 mol H2(g)? Assume that N2(g) is the excess reactant.
Answer:
22 g
Explanation:
Step 1: Write the balanced equation
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
Step 2: Calculate the moles of NH₃ produced from 2.0 moles of H₂
The molar ratio of H₂ to NH₃ is 3:2.
2.0 mol H₂ × 2 mol NH₃/3 mol H₂ = 1.3 mol NH₃
Step 3: Calculate the mass corresponding to 1.3 moles of NH₃
The molar mass of NH₃ is 17.03 g/mol.
1.3 mol × 17.03 g/mol = 22 g
Why are watersheds important for the maintenance of water quality? Choose all that apply.
Select 2 correct answer(s)
Question 3 options:
they can transfer pollutants
their soil and plants can filter and clean water
they cause precipitation and acid rain
they cause floods
Answer:
its A and D
Explanation:
i took the quiz
Answer:
a and d
Explanation:
A radio wave travels 186,000 miles per second. How many kilometers will the wave travel in one microsecond? (1 mile = 1.61 km)
need steps and formela
How many grams are in 2.49 x 10^24 atoms of Hg?
Answer:
[tex]m_{Hg}=829.4gHg[/tex]
Explanation:
Hello there!
In this case, considering the Avogadro's number, which helps us to realize that 1 mole of mercury atoms contains 6.022x10²³ atoms and at the same time 1 mole of mercury weights 200.59 g, we obtain:
[tex]m_{Hg}=2.49x10^{24}atoms*\frac{1mol}{6.022x10^{23}atoms} *\frac{200.59g}{1molHg}\\\\m_{Hg}=829.4gHg[/tex]
Best regards!
If you begin with 14.0g of TiCl4(1), how many liters of H2O(g) will need?
Answer:
3.32 L
Explanation:
Step 1: Write the balanced equation
TiCl₄(l) + 2 H₂O(g) → TiO₂(s) + 4 HCl(g)
Step 2: Calculate the moles corresponding to 14.0 g of TiCl₄
The molar mass of TiCl₄ is 189.68 g/mol.
14.0 g × 1 mol/189.68 g = 0.0738 mol
Step 3: Calculate the moles of H₂O needed to react with 0.0738 moles of TiCl₄
The molar ratio of TiCl₄ to H₂O is 1:2. The moles of H₂O needed are 2/1 × 0.0738 mol = 0.148 mol
Step 4: Calculate the volume corresponding to 0.148 moles of H₂O(g)
At standard temperature and pressure, 1 mole of H₂O(g) has a volume of 22.4 L.
0.148 mol × 22.4 L/1 mol = 3.32 L
Two experiments were conducted: 4.125 g of copper(II) oxide, CuO2, was reduced by heating at high pressure with excess hydrogen to yield 3.294 g of copper Cu. 1.179 g of copper Cu was dissolved in nitric acid to yield copper(II) nitrate, which was converted to 1.476 g copper(II) oxide, CuO2, on ignition. Show that the results illustrate the law of definite proportions by calculating the percentage of oxygen in copper(II) oxide in experiments 1 and 2.
Answer:
Explanation:
4.125 g of copper oxide contains 3.294 g of Cu . Rest is oxygen
oxygen is 4.125 - 3.294 = .831 g
Ratio of oxygen and copper = .831 : 3.294
.831 / .831 : 3.294 / .831
= 1 : 3.96
1 : 4 ( Approx )
percentage of oxygen in copper oxide = 20 % approx
in the second experiment ,
1.476 g of copper oxide is formed from 1.179 g of Cu
oxygen in this copper oxide = 1.476 - 1.179 = .297 g
Ratio of oxygen and copper
.297 : 1.179
.297/.297 : 1.179/ .297
1 : 3.96
1 : 4 ( approx )
percentage of oxygen in copper oxide = 20 % approx
So it follows the law of definite proportions .
Determine the number of moles of air present in 1.35 L at 750 torr and 17.0°C. Which equation should you use? COMPLETE What is the number of moles present?
Answer:
0.0559 mol
Explanation:
Step 1: Given and required data
Volume of air (V): 1.35 LPressure of air (P): 750 torrTemperature (T): 17.0 °CIdeal gas constant (R): 0.0821 atm.L/mol.KStep 2: Convert "P" to atm
We will use the conversion factor 1 atm = 760 torr.
750 torr × 1 atm/760 torr = 0.987 atm
Step 3: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 17.0 °C + 273.15 = 290.2 K
Step 4: Calculate the number of moles of air
If we assume air behaves as an ideal gas, we can calculate the moles (n) of air using the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.987 atm × 1.35 L/(0.0821 atm.L/mol.K) × 290.2 K = 0.0559 mol
The most prominent feature of ocean topography discovered in the 1960s was:
a mountain ridge which circles the earth
subduction zones
volcanic activity at the ocean bottom
seamounts
answer:
a mountain ridge which circles the earth
subduction zones
How many protons, neutrons, and electrons are in the isotope 32S2-
Answer:
There are 16 protons, 18 electrons and 16 neutrons. I hope this helped:)
short note on dung fertilization
Answer:
Most manure consists of animal feces; other sources include compost and green manure. Manures contribute to the fertility of soil by adding organic matter and nutrients, such as nitrogen, that are utilised by bacteria, fungi and other organisms in the soil.
Explanation:
Hope it is helpful....
Answer:
Animal manure, such as chicken manure and cow dung, has been used for centuries as a fertilizer for farming. It can improve the soil structure (aggregation) so that the soil holds more nutrients and water, and therefore becomes more fertile. ... Manure is also commercially composted and bagged and sold as a soil amendment.
Explanation:
Timmy pushes off of the pool wall with a force of 2300 N and accelerates at 15 m/
what is mass? (ignore water resistance)
use one decimal place in your answer
Please help!
Answer:
Mass = 153.33 Kg
Explanation:
Given:
Force = 2,300 N
Acceleration = 15 m/s²
Find:
Mass
Computation:
Mass = Force / Acceleration
Mass = 2,300 / 15
Mass = 153.33 Kg
write the balanced chemical equation for lab preparation of oxygen gas by using heat
Answer:
O + O ---Δ----> O2
Explanation:
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Answer:
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Answer the following question:
How many moles of ammonia are in 375. mL of a 2.25 M aqueous ammonia solution?
Find the numerical answer for this question and make sure to include the following:
What is the chemical formula for ammonia?
What is the formula for molarity?
When you give your numerical answer, what is the correct significant figures and how do you know that is the correct amount?
Answer:
NH₃
M = n/V(L)
0.844 mol (Both numbers have 3 significant figures so the result has 3 significant figures as well)
Explanation:
Step 1: Given and required data
Volume of solution (V): 375. mLMolar concentration of the solution (M): 2.25 MChemical formula for ammonia: NH₃Step 2: Calculate the moles (n) of ammonia (solute)
Molarity is equal to the moles of solute divided by the liters of solution.
M = n/V(L)
n = M × V(L)
n = 2.25 mol/L × 0.375 L = 0.844 mol (Both numbers have 3 significant figures so the result has 3 significant figures as well)
(trying this again because this test is due at 3 and paimon has to get this done or she will fail!! paimon will try to give brainlist if brainly lets her!!)
The _________________molecules in the food we eat are_____________.
A. Enzyme, water
B. Small, insoluble
C. Large, insoluble
D. Enzyme, insoluble