Answer:
30 yard per second
Explanation:
300/10 = 30
The night sky is filled with images of wild animals. Which animal is the circled constellation named after?
scorpion
serpent
snake
salamander
A group of stars connected by lines is circled and labeled Saturn.
Answer:
ccc
Explanation:
Answer:
A
Explanation:
trust
Two units of distance used by scientists to describe distances in space are the light-year and astronomical unit. Why do scientists use the light-year instead of Astronomical Units to measure the distances between stars?
The speed of sound is too slow.
Telescopes can measure only in light years
Distant stars can only be seen if there is light
The smaller numbers that result from measuring with larger units are easier to use
Answer:
The smaller numbers that result from measuring with larger units are easier to use .
That’s the one, good luck!
The smaller numbers that result from measuring with larger units are easier to use.
Light year is a unit of astronomical distance that measures the distance traveled by light in one year. The distance between stars can also be measured using parsecs.
The equivalent of this distance in various units is written as follows;
1 light year (ly) = 9.5 x 10¹² km = 63,240 AU1 parsec = 3.26 lyUsing light year to represent the unit of distance is more concise because big measurement can be represented easily with small numbers using this large unit.
Thus, we can conclude that the smaller numbers that result from measuring with larger units are easier to use.
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Predict the chemical formula of the compound that will form when Li and o interact.
Answer:
[tex]Li_2O[/tex]
Explanation:
Hello!
In this case, when looking over the chemical formula of a compound formed when two elements interact, we need to realize about their location in the periodic table; thus, since lithium is a metal and oxygen a nonmetal, we infer lithium turns out the cation and oxygen the anion. Moreover, as the oxidation state of lithium is 1+ and that of oxygen is 2-, we can set up the chemical formula as follows:
[tex]Li^+O^{2-}\\\\Li_2O[/tex]
Best regards!
If you had the same stock of the blue solution as in the serial dilution simulation you did (1M solution), and you needed 10 mL of a solution that was 1.0x10-6 M, what volume of the blue stock solution would you need, and what volume of diluent would you need to reach 10 mL of the desired concentration
Answer: We start by doing a 1/10 serial dilution, using 100 µL of 1M solution into 900 µL of water, until we get a 1.0x10^-5M (0.00001M) solution. Then use 1 mL of this 1.0x10^-5M solution into 9 mL of water.
Explanation:
To answer this question, we must use the law of conservation of mass, which states: In every chemical reaction mass is conserved, this means the total mass of the reactants is equal to the total mass of the products. The law implies that mass can neither be created nor destroyed, but it can be transformed. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction. Therefore, during any chemical reaction and low-energy thermodynamic processes in an isolated system, the total mass of the reactants or starting materials must be equal to the mass of the products. This law is quite accurate for low-energy processes, such as chemical reactions.
So, if the solution to be used is 1M , and a 10 mL of a 1.0x10^-6M solution is needed, we use the following equation:
Initial concentration x initial volume = final concentration x final volume.
The initial concentration is 1M, the final concentration is 1.0x10^-6M and the final volume required is 10 mL.
1M x initial volume = 1.0x10^-6M x 10 mL
initial volume= 1.0x10^-5 mL= 0.01 µL of 1M solution.
Since the final volume is 10 mL, we have to add the difference in volume with water, which is 10 mL - 1.0x10^-5 mL= 9.99999 mL.
However, since 0.01 mL is a very small volume that is difficult to take, the best option in this case is to make serial dilutions.
Usually, we start from a concentrated solution and prepare a series of dilutions to the tenth (1:10) or half (1:2). In this way a series of solutions is obtained, related for example by a dilution factor of 10, i.e. 1/10; 1/100; 1/1000 and so on.
Here we can prepare a series of dilutions to the tenth, from 1M to 1.0x10^-6.
We start by doing a 1/10 dilution, using 100 µL of 1M solution into 900 µL of water. This is a 0.1M solution. Then we take 100 µL of it into 900 µL of water to get a 0.01M solution. We continue doing that until we get a 1.0x10^-6M (0.000001M) solution. This final solution is the desired concentration, however we need 10 mL of it, and actually we have 1 mL. So we can just take 1 mL of the 1.0x10^-5M solution into 9 mL of water:
1.0x10^-5M x 1 mL = 1.0x10^-6M x 10 mL.
Which of the following observations about burning sugar provides evidence of a
chemical reaction?
a. Heat is added to the sugar crystals.
b. The sugar melts and becomes a liquid.
с
C. The temperature of the sugar increases.
d. Gas is produced as the sugar turns black.
1 Pt
А
B
D
Please help asap! Giving brainliest to correct answer.
Three sections of the periodic table are labeled A, B, and C in the image below.
Which of the following statements is most likely true for an element present in section B?
Select one:
a. It is a highly reactive metal.
b. It is an inert gas at room temperature.
c. It may act as an insulator or a conductor.
d. It is a highly reactive liquid at room temperature.
Answer:
C. It may act as an insulator or a conductor.
Explanation:
Group B are the metalloids, so they have properties of both metals and nonmetals.
The statement which is most likely true for an element present in section B is: C. It may act as an insulator or a conductor.
A periodic table can be defined as an organized tabular array of all the chemical elements, which is typically arranged in order of increasing atomic number or number of protons (in rows).
In this diagram, the chemical element that is present in section B of the periodic table is a metalloid.
A metalloid can be defined as a chemical element that are typically intermediate between those of semiconductors, non-metals and metals.
Thus, a metalloid has both chemical and physical properties that are a combination of metals and non-metals or semiconductors.In this context, the statement which is most likely true for an element present in section B is that it may act as an insulator or a conductor because it is a metalloid.
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What makes an atom a good conductor?
If 80% of the moths are dark and you have 80 moths, how
many would be dark??
Answer:
the answer is 64 moths are dark
Help pls!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
water/d
Explanation:
Which of the following are made possible by energy transfer from the Sun? Choose all that apply.
A. biofuels
B. nuclear power using uranium
C. wind power
D. hydroelectric power using a dam
This is Really Science but I couldn’t find it but please help ASAP!
g) How many grams of aluminum (Al) are present in 2.98 mol of Al?
h) How many moles of Br are there if the sample contained 5.6 x 1024 atoms of Br?
Answer:
26.982 g
i think so..........
Which of these is smallest?
cells
atoms
matter
elements
Answer:
atoms
Explanation:
i hope that help you
Answer:
atoms
Explanation:
How many moles are in 3.30 x 1024 molecules of N2I6
Divide the number of molecules N by avogadro's constant.
Let n be the number of moles then,
[tex]n=\frac{N}{N_A}=\frac{3.3\cdot10^{24}}{6.022\cdot10^{23}}=5.48\mathrm{mol}[/tex]
Hope this helps.
Which process directly moves nutrients from plants to the air
A) Percolation
B) Transportation
C) Assimilation
D) Photosynthesis
Answer:
Plants release oxygen in the process of Photosynthesis
Explanation:
What does "X" represent in the following symbol?
80
Х
35
Answer:bromine
Explanation:
what is this for questions 2
Answer:
I think is D
Explanation:
PLEASE MARK ME AS BRAINLIEST
How many atoms are found in a 15.5 g sample of bismuth?
Answer:
The number of particles in the 41.8g sample of Bismuth is 12.044 × 10²³
Explanation:
In a 15.5 g sample of bismuth there are 4.47 x 10²² atoms are present.
What do you mean by the term molar mass ?The term molar mass is defined as the ratio between the mass and the amount of substance of any sample of said compound. The molar mass is a property of a substance.
The molar mass of a compound can be calculated by adding the standard atomic masses in g/mol of the constituent atoms.
Given:
Mass of Bismuth = 15.5g
Number of atoms = ?
To find the number atoms, find the number of moles in this element first.
Number of moles = mass/ molar mass
Number of moles = 15.5g / 209
= 0.074mole
1 mole of a substance contains 6.023 x 10²³ atoms
0.074 mole of Bismuth will contain 0.074 x 6.023 x 10²³ atoms
= 4.47 x 10²² atoms
Thus, 4.47 x 10²² atoms are found in a 15.5 g sample of bismuth.
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According to the following reaction, how many moles of calcium chloride will be formed upon the complete reaction of 27.9 grams of calcium hydroxide with excess hydrochloric acid? calcium hydroxide (aq) + hydrochloric acid (aq) calcium chloride (aq) + water (l) moles calcium chloride
Answer: 0.377 moles of calcium chloride will be formed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Ca(OH)_2=\frac{27.9g}{74g/mol}=0.377moles[/tex]
The balanced chemical reaction is:
[tex]Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O[/tex]
According to stoichiometry :
As 1 mole of [tex]Ca(OH)_2[/tex] give = 1 mole of [tex]CaCl_2[/tex]
Thus 0.377 moles of [tex]Ca(OH)_2[/tex] give =[tex]\frac{1}{1}\times 0.377=0.377moles[/tex] of [tex]CaCl_2[/tex]
Thus 0.377 moles of calcium chloride will be formed upon the complete reaction of 27.9 grams of calcium hydroxide with excess hydrochloric acid
what happens when hydrogen gas is passed over hot ferric oxide plzz help
When hydrogen is passed over hot ferric oxide (FeO) hydrogen reacts with oxygen present in the compound and forms water (H2O) and pure Iron
compute the mass-specific enthalpy change associated with Nz that is undergoing a change in state from 400 k to 800 k
Answer:
The correct answer is "430 kJ/kg". A further explanation is given below.
Explanation:
The given values are:
T₁ = 400 k
T₂ = 800 k
The average temperature will be:
= [tex]\frac{T_1+T_2}{2}[/tex]
= [tex]\frac{400+800}{2}[/tex]
= [tex]600 \ k[/tex]
From table,
At 600 k the [tex]C p[/tex] will be = 1.075
Now,
⇒ The specific enthalpy = [tex]Cp(T_2-T_1)[/tex]
⇒ [tex]\Delta h=1.075 (800-700)[/tex]
⇒ [tex]=430 \ kJ/kg[/tex]
It’s a test and only need 2 more problems!!!! Thank you and Appreciate it!!!
Answer:
the molar mass of propane (c3H8) is
12*3+1*8
=36+8=44
15 The picture shows Earth’s rotation. Which pattern listed below is caused by Earth’s rotation? A Water cycle processes B Length of shadows C Seasonal temperatures D Weathering and erosion HELP ME
Answer:
jayfeather friend me
Explanation:
Hence it is quite important to understand and learn the processes of the water cycle.
Step 1: Evaporation. The water cycle starts with evaporation. ...
Step 2: Condensation. ...
Step 3: Sublimation. ...
Step 4: Precipitation. ...
Step 5: Transpiration. ...
Step 6: Runoff. ...
Step 7: Infiltration.
Will give brainliest to correct answer of this picture:
Answer:
option C : cellular respiration
Explanation:
cellular respiration is taking place in this diagram .
as in respiration food get oxidised to release energy and it releases CO2 and water
Don’t skip please!!! I need help
Given
Atomic symbol
Required
Atomic composition
Solution
Atomic number = number of protons = number of electrons
Mass Number (A) is the sum of protons and neutrons
Number of protons and Number of Neutrons in nucleus
Number of electrons in the shell
From the picture:
protons = 3
neutrons = 4
electrons = 3
atomic number = protons = electrons = 3
mass number = protons + neutrons = 3+4 = 7
What is the volume in mL of 0.36 mol of carbon dioxide at STP?
Answer:
Zero
Explanation:
The reaction A→B is to be carried out isothermally in a continuous-flow reactor. Calculate the PFR volume to consume 99% of A (CA = 0.01CA0) when the entering molar flow rate is 5 mol/h (assume pure A), the volumetric flow rate is constant at 10 dm3 /h and the rate is: -rA=3CA 2 [ dm3 /mol•h]
Answer:
Explanation:
From the information above:
The equation for the reactor volume of CSTR is given by the formula:
[tex]V = \dfrac{C_{Ao}V_o - C_A V_o }{-rA}[/tex]
[tex]V = \dfrac{C_{Ao}V_o (1 - 0.01)}{KC_A^2}[/tex]
The reaction rate of A is:
[tex]-r_A = kC_A^2[/tex] and k = 3 dm³/mol.h
replacing the values in the above equation to solve for v;
∴
[tex]V = \dfrac{(0.5 \ mol/dm^3) (10 \ dm^3/h) (1 - 0.01)}{ ( 3 \ dm^3/mol.h) (0.01 \times 0.5 \ mol/dm^3)^2}[/tex]
V = 66,000 dm³
Thus, the reactor volume of CSTR is 66000 dm³
Taking the differential mole balance equation for PFR is as follows:
[tex]\dfrac{d(C_AV_o)}{dV} = -r_A[/tex]
[tex]\dfrac{d(C_AV_o)}{dV} = Kc^2A[/tex]
Taking the integral between initial and final concentrations;
[tex]\dfrac{v_o}{k} \int ^{C_A}_{C_{Ao}} \dfrac{dC_A}{C^2A}= - \int ^V_o \ dV[/tex]
[tex]V = \dfrac{V_o}{K} \Big [\dfrac{1}{C_A}- \dfrac{1}{C_{Ao}} \Big ][/tex]
replace the values in the above equation;
[tex]V= \dfrac{(10 \ dm^3/h) }{(3 \ dm^3/mol.h)}\Big [ \dfrac{1}{0.01(0.5 \ mol/dm^3)} - \dfrac{1}{0.5 \ mol/dm^3} \Big ][/tex]
[tex]\mathbf{V = 66 0 \ dm^3}[/tex]
Thus, the reactor volume of PFR is 660 dm³
Determine the mass (in g) of a substance, with a density of 2.26 g/mL, that occupies 350.0 mL.
Answer:
m = 79.1 grams
Explanation:
Given that,
Density, d = 2.26 g/mL
Volume occupies, V = 35 mL
We need to find the mass of the substance. We know that the density of an object is equal to the mass per unit volume. So,
d = m/V
[tex]m=d\times V\\\\m=2.26\ g/mL\times 35\ mL\\\\m=79.1\ g[/tex]
So, the mass of the substance is 79.1 grams.
Question 12 (1 point)
You notice light pollution in
a.space
b.big cities
c.rural areas
d.our galaxy
Answer:
the answer is D . our galaxy
Answer:
our galaxy of course
One way to represent this equilibrium is: 2 Al(s) 3 Br2(l)2 AlBr3(s) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.
The question is incomplete, the complete question is;
Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:
Al(s) + 3/2 Br2(l)AlBr3(s)
We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.
1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)
2) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)
3) AlBr3(s) Al(s) + 3/2 Br2(l)
Answer:
See explanation
Explanation:
We have that; Al(s) + 3/2 Br2(l)AlBr3(s)
So;
Al(s) + 3/2 Br₂(l) = AlBr₃(s)
K = [ AlBr₃] / [ Al] [ Br₂]³/²
K² = [ AlBr₃]² / [ Al ] ² [ Br₂]³
Now;
1) 2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =
K₁ = [ Al ] ² [ Br₂]³ / [ AlBr₃]²
K₁ = ( 1 / K² ) = K⁻²
For the second reaction;
2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)
K₂ = [ AlBr₃ ]² / [ Al ]² [ Br₂ ]³
K₂ = K²
For the third reaction;
3 )
AlBr₃(s) = Al(s) + 3/2 Br₂(l)
K₃ = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]
= ( 1 / K ) = K⁻¹
[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was conducted on the soil by allowing the sample to drain for 24 hours. After drainage the sample mass was 295 g. The soil was then dried and weighed 284 g. What are the specific yield [8 points], specific retention [8 points], and porosity [8 points] of the sample
Answer:
Follows are the solution to the given question:
Explanation:
Dry Soil weight = solid soil weight = [tex]284 \ grams[/tex]
solid soil volume =[tex]205 \ cc[/tex]
saturated mass soil = [tex]361 \ g[/tex]
The weight of the soil after drainage is =[tex]295 \ g[/tex]
Water weight for soil saturation = [tex](361-284) = 77 \ g[/tex]
Water volume required for soil saturation =[tex]\frac{77}{1} = 77 \ cc[/tex]
Sample volume of water: [tex]= \frac{\text{water density}}{\text{water density input}}[/tex]
[tex]= 361- 295 \\\\ = 66 \ cc[/tex]
Soil water retained volume = (draining field weight - dry soil weight)
[tex]= 295 - 284 \\\\ = 11 \ cc.[/tex]
[tex]\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}[/tex]
[tex]= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%[/tex]
(Its saturated water volume is equal to the volume of voids)
[tex]\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}[/tex]
[tex]= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23[/tex]
[tex]\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}[/tex]
[tex]= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04[/tex]
