500kg, 42m, 675g, and 10m/s are all examples of a: continuous data.
A numerical data can be defined as a data set that is expressed in numbers only or a data set consisting of numbers rather than words. A numerical data is also known as a quantitative data.
Generally, numerical data are classified into two (2) main categories and these are;
Discrete data: a discrete data is a data set in which the number of possible values are either finite or countable. Some examples are; the value of a fair die, number of pen in a box, number of eggs in a crate, etc.Continuous data: a continuous data is a data set that has infinitely many possible values and as such those values cannot be counted, which simply means they are uncountable. Any data (quantity) such as mass, height, volume, weight, density, velocity, length, pressure, temperature, speed, distance, time are generally classified as continuous data.In conclusion, mass (500 kg), height (42m), mass (675g), and speed (10m/s) are all examples of a: continuous data.
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What does empirical fomula means with examples
Can someone tell me the advantages and disadvantages of Ceramics? I need scientific, good and clear points! :)
what is the correct electron configuration for an element with five electrons in the third energy level
[tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex] is the correct electron configuration for an element with five electrons in the third energy level.
What are elements?Elements are the simplest substances which cannot be broken down using chemical methods.
The shell nearest to the nucleus, 1n, can carry two electrons, while the next shell, 2n, can carry eight, and the third shell, 3n, can carry up to eighteen.
The third shell carries 18 electrons; 2 in a 3s orbital; 6 in three 3p orbitals; and 10 in five 3d orbitals. The fourth shell carries 32 electrons; 2 in a 4s orbital; 6 in three 4p orbitals; 10 in five 4d orbitals; and 14 in seven 4f orbitals.
The element would be phosphorus. Its electron configuration is [tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex]
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What is 16.30 to the 100th gram
Answer:
1.65485688 × 10121 grams
Explanation:
Aside from those discrete items, anything like rice or flour can be measured out to make up 100 grams. For example, 1/2 cup (125 ml or about 4 and 1/2 fl. oz) of dry rice weighs very close to 100 grams.
26. Calculate the mass of hydrogen formed when 25 grams of aluminum reacts with excess hydrochloride acid. 2AI + 6HCl →l Al2Cl6 + 3H2
Answer:
2.76 gram
Explanation:
2AI + 6HCl → 2AlCl3 + 3H2
get the mol of aluminum => n = m / M = 25 / 27 = 0.925 (mol)get the ratio mol of Hydro => nH2 = 0.925 * 3 / 2 = 1.38 (mol)get weight of hydro mH2 = n x M = 1.38 x 2 = 2.76 (gram)Answer
Al +2 HCl - AlCl2 + H2
Which statement defines an ore?
Answer:
The third one
Explanation:
An ore is a naturally occurring solid material from which a metal or valuable mineral can be profitably extracted.
boron has two naturally occurring isotopes: boron -10 (abundance = 19.8% , mass = 10.013 amu) boron -11 (abundance = 80.2%, mass= 11.009 amu) Calculate th eatomic mass of boron
Explanation:
eam=%abudance×mass+%abudance×mass
eam=19.8%×10.013/100+80.2%×11.009/100
eam=198.2574/100+882.9218/100
eam=1081.1792/100
eam=10.811792
eam=10.812
Plsss help with this ASAP TT
Answer: lithium- 3 protons, 3 electrons, 3 neutrons
Oxygen- nucleon no. 16, Atomic no. 8, 8 protons, overall charge (-2)
Sorry I don't the others. Hope this helps
Chromium is dissolved in sulfuric acid according to the following equation: Cr + H2SO4 ⇒ Cr2 (SO4) 3 + H2
a) How many grams of Cr2 (SO4) 3 can be obtained by reacting 165 g of 85.67% H2SO4 of purity?
b) If 485.9 g of Cr2 (SO4) 3 are obtained, what is the yield of the reaction?
Answer:
[tex]\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}[/tex]
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 98.08 392.18
2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂
To solve the stoichiometry problem, you must
Use the molar mass of H₂SO₄ to convert the mass of H₂SO₄ to moles of H₂SO₄ Use the molar ratio to convert moles of H₂SO₄ to moles of Cr₂(SO₄)₃ Use the molar mass of Cr₂(SO₄)₃ to convert moles of Cr₂(SO₄)₃ to mass of Cr₂(SO₄)₃
a) Mass of Cr₂(SO₄)₃
(i) Mass of pure H₂SO₄
[tex]\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}[/tex]
(ii) Moles of H₂SO₄
[tex]\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}[/tex]
(iii) Moles of Cr₂(SO₄)₃
The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ [tex]\text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}[/tex]
(iv) Mass of Cr₂(SO₄)₃ [tex]\text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}[/tex]
b) Percentage yield
It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.
[tex]\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}[/tex]
Cubes are three-dimensional square shapes that have equal sides. What is the density of a cube that has a mass of 12.6 g and a measured side length of 4.1 cm? (Density: D = )
Answer:
0.1828g/cm³
Explanation:
density= mass÷volume
m= 12.6g
v= 4.1×4.1×4.1 = 68.921cm³
•
density= 12.6÷68.921 = 0.1828g/cm³
what is the mass of 1 mole of chlorine ion
Answer:
35.453 g
Explanation:
one mole of chlorine is 34.453 g
Answer:
One mole of chlorine atoms has a mass of 35.453g. A mole of something means that you have 6.023×1023 units of that something. So if you have a mole of Cl atoms you have 6.023×10
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Calculate the pressure drop over a 100m length due to friction when a slurry made from 1.0-mm silica particles is pumped through a horizontal 6-cm diameter pipeline (smooth pipe) at 2.5 m/s. The slurry contains 25 per cent silica by volume. The density of silica is 2700 kg/m3, rhow = 1000 kg/m3, μw = 0.001 kg/ms. Use a value of 82 for Ω. Drag coefficient for these particles,CD, may be taken as 0.44 .
For the slurry/Pipe system in question 1, estimate the deposition velocity.
Answer:
The answer is "2.78".
Explanation:
Given values:
CD= 0.44
Formula:
[tex]\bold{f_r= \frac{v^{2}}{ g(s-1)D}}[/tex]
g=9.8
s= 2.7
D= 0.06
[tex]\to f_r=\frac{2.5^2}{9.8(2.7-1)0.06}\\\\[/tex]
[tex]=\frac{2.5 \times 2.5}{9.8 \times 1.7 \times 0.06 }\\\\=\frac{6.25}{.9996 }\\\\=6.252501[/tex]
[tex]\phi = \frac{82\times v}{ \sqrt{cD} \times f_r^{-1.5}}\\\\[/tex]
[tex]= \frac{82 \times 0.25 }{ \sqrt{0.44} \times 6.25^{1.5}}\\\\=2.4285\\[/tex]
[tex]\frac{\bigtriangleup P f_1 s_1}{L} = \frac{\bigtriangleup Pf_w}{L}(1+\phi)\\[/tex]
[tex]=\frac{2fwSwv^2 (1+2.4285)}{D}\\\\[/tex]
[tex]Re= \frac{D \bar v Sw}{M_w}\\[/tex]
[tex]=\frac{0.06 \times 2.5 \times 1000 }{0.001}\\\\=\frac{150 }{0.001}\\\\= 150 \times 10^{3}\\\\= 1.50 \times 10^{5}\\\\[/tex]
[tex]fw= 0.00389[/tex]
[tex]\to \frac{\bigtriangleup P f_1 s_1}{L}[/tex]
[tex]\to \frac{2 \times 0.00389 \times 1000 \times2.5^2 \times 3.4265}{0.06}\\\\\to 2.78[/tex]