Answer:
40.kPa hope this helps....
Please help asap. Brainliest to correct.
Which formula can be used to calculate the percent yield? (5 points)
Select one:
a. (Theoretical yield ÷ actual yield) x 100
b. (Actual yield ÷ theoretical yield) x 100
c. (Amount of reactants ÷ amount of products) x 100
d. (Amount of products ÷ amount of reactants) x 100
Answer:
I guess this answer is c
yh
Answer:
b. (Actual yield ÷ theoretical yield) x 100
Explanation:
The percent yield will be a ratio of collected or actual yield to the theoretical yield. Like any percentage, you are determining a part of a whole. You multiply by 100 to convert your decimal answer from the division into a percentage.
4. Describe a simple experiment that can be performed in the laboratory to demonstrate
the formation of iron (III) chloride from iron fillings.
Answer:
laboratory use the organic substances dissolve in water methanol or ethanol then Sinha to rise iron chloride solution is added a transient permanent coloration usually purple green or blue indicate the presence of a funnel or an hall
Balanced equation for Fe reacting with [tex]Cl_2[/tex]
[tex]2Fe + 3Cl_2[/tex] → [tex]2FeCl_3[/tex]
What are iron fillings?Iron filings are small shavings of ferromagnetic material.
When hydrochloric acid (HCl) reacts with iron filings (Fe) a dissolving metal reaction occurs with the Fe being converted to iron chloride ([tex]FeCl_2[/tex]) and hydrogen ([tex]H_2[/tex]).
The balanced equation is:
[tex]Fe + 2HCl[/tex] → [tex]FeCl_2 + H_2[/tex]
Note that reaction only produces [tex]FeCl_2[/tex] and not [tex]FeCl_3[/tex]. To make [tex]FeCl_3[/tex] you have to get more forcing and react Fe with chlorine gas ([tex]Cl_2[/tex]) at elevated temperature.
Since we are balancing equations here, below is the balanced equation for Fe reacting with [tex]Cl_2[/tex] (offered at no extra charge):
[tex]2Fe + 3Cl_2[/tex] → [tex]2FeCl_3[/tex]
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Is the bond length in HCl the same as that in DCl? The wavenumbers of the J = 0 1 rotational transitions for H35Cl and 2H35Cl are 20.8784 and 10.7840 cm–1, respectively. Accurate atomic masses are H = 1.007825 amu, 2H = 2.0140 amu, and 35Cl = 34.96885 amu. Based on this information alone, can you conclude that the bond lengths are the same or different in the two molecules?
how many lone pairs are on the central atom in GeH4?
Is this statement true or false?
All gymnosperms lose their leaves in the winter.
Answer:
they do fall off of themselves
Answer:
True!
Explanation:
They usually fall off in the winter
LEEN was observing the physical properties of elements in order to classify them as either a metal, non-metal, metalloid or any other type of element. The first element was dull, brittle, and did not conduct heat. What would its classification be
Metal
Non-metal
Metalloid
Radioactive element
Answer:
Non-metal
Explanation:
A non-metal is a substance that possesses the following characteristics:
- It is brittle i.e can break easily
- It does not conduct heat and electricity
- It is not lustrous i.e does not shine when polished.
According to this question, LEEN was observing the physical properties of elements in order to classify them. He found that the first element was dull, brittle, and did not conduct heat. This makes it a NON-METAL.
electronic configuration [Ne]3s2 3p6
calculate the wavelength of the first line in the lymen series of hydrogen spectrum
Plz help Need Answer Quick
Answer:
The correct answer is - [tex]\frac{P}{4}[/tex].
Explanation:
By the ideal gas equation is PV=nRT, the relation between pressure and volume can be drive if the temperature remains constant:
At constant temperature, for a fixed amount of gas, P∝
so the pressure is inversely proportional to volume. Therefore, if the volume is increased in this condition the pressure will decrease in the same manner and A decrease in the volume will lead to an increase in pressure.
Volume increased here = four times = 4V
pressure is P will decrease four times = P/4
What is the mass of 5.55 moles of carbon monoxide
Answer:
[tex]\boxed {\boxed {\sf 155 \ g\ CO}}[/tex]
Explanation:
To convert from moles to mass, the molar mass must be used.
First, write the chemical formula for carbon monoxide. Since the carbon (C) comes first without a prefix, there is 1 carbon atom. The prefix mono- before oxide means 1, so there is also 1 oxygen (O) atom. The formula is CO.
Next, look up their molar masses on the Periodic Table.
C: 12.011 g/mol O: 15.999 g/molSince there is 1 atom of each, the molar masses can be added.
CO: 12.011 g/mol + 15.999 g/mol = 28.01 g/molUse this molar mass as a ratio.
[tex]\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]
Multiply by the given number of moles:5.55
[tex]5.55 \ mol \ CO *\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]
The moles of carbon monoxide cancel.
[tex]5.55 * \frac {28.01 \ g \ CO} {1 }[/tex]
Multiply.
[tex]155.4555 \ g \ CO[/tex]
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, it is the ones place. The 4 in the tenths place tells us to leave the 5.
[tex]155 \ g\ CO[/tex]
5.55 moles of carbon monoxide is about 155 grams.
The total pressure in a container is 37.9 psi. There are two gases in the container. Gas A exerts a pressure of 18.4 psi. What is the pressure of gas B?
Answer:
19.5
Explanation: using dalton law pt=p1+p2+p3...
the total pressure is 37.9 so to get pressure of gas b subtract pressure of gas a from total pressure.37.9-18.4 gas b equals 19.5
The lithosphere ______ part of the earth’s crust.
A. Gaseous
B. Solid
C. Water
D. Liquid
Answer:
The lithosphere solid part of the earth's crust
Methane is combusted with Oxygen to produce Carbon Dioxide and Water. How many grams of oxygen are required to react with 55 grams of
Methane?
A.55 g
B.110 g
C.165 g
D.220 g
2). H2S and SO2 as follows,(8pt)
H2S + SO2 ------>S + H2O
In a particular experiment, 7.95 g of sulfur is produced by the reaction of 7.5 g of H2S with 12.75 g of SO2.
What is the % yield of sulfur?
Answer: 75.7 % yield
Explanation:
The balanced chemical equation is:
[tex]2H_2S+SO_2\rightarrow 3S+2H_2O[/tex]
According to stoichiometry :
68.2 g of [tex]H_2S[/tex] will require = 64 g of [tex]SO_2[/tex]
Thus 7.5 g of [tex]H_2S[/tex] will require = [tex]\frac{64}{68.2}\times 7.5=7.0g[/tex] of [tex]SO_2[/tex]
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 68.2 g of give = 96 g of
Thus 7.5 g of [tex]H_2S[/tex] give =[tex]\frac{96}{68.2}\times 7.5=10.5g[/tex] of [tex]S[/tex]
% yield=[tex]\frac{\text{Actual yield}}{\text {Theoretical yield}}\times 100=\frac{7.95g}{10.5g}\times 100=75.7\%[/tex]
Thus 75.7 % yield is there.
Which carbohydrate strengthens plant cells?
A. glucose
B. glycogen
C. cellulose
D. lactose
Answer:
a
Explanation:
Answer:
c
Explanation:
edge
calculate the percentage of sodium and carbon and oxygen in sodium carbonate
What cell tissue helps carry messages? :
options:
muscular
nervous
connective
Answer:
Nervous
Explanation:
Nervous tissue is composed of two main cell types: neurons and glial cells. Neurons transmit nerve messages.
2 NaOH + H2SO4 → 2 H2O + Na2SO4
How many grams of water are produced from 203.50 grams of H2S04?
Answer:
[tex]m_{H_2O}=74.8gH_2O[/tex]
Explanation:
Hello there!
In this case, according to the given chemical reaction, it is possible to realize there is a 1:2 mole ratio of sulfuric acid to water; thus, given the mass of the former and its molar mass (98.07 g/mol), it is possible to determine the mass of produced water as shown below:
[tex]m_{H_2O}=203.50gH_2SO_4*\frac{1molH_2SO_4}{98.07gH_2SO_4}*\frac{2molH_2O}{1molH_2SO_4} *\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=74.8gH_2O[/tex]
Regards!
If a ball rolling down a hill is half way between the top and bottom, how much potential energy does the ball have compared to kinetic energy?
Answer:
The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)
Explanation:
The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.
If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.
If the gravitational field strength [tex]g[/tex] is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.
If [tex]m[/tex] denote the mass of this ball, the gravitational potential energy ([tex]\rm GPE[/tex]) of this ball at height [tex]h[/tex] would be [tex]{\rm GPE} = (m \cdot g) \cdot h[/tex], which is proportional to [tex]h\![/tex].
The value of [tex]g[/tex] near the surface of the earth is indeed approximately constant (typically [tex]g \approx 9.8\; \rm m \cdot s^{-2}[/tex].)
At halfway between the top and bottom of this hill, the height of this ball would be [tex](1/2)[/tex] of its initial value (the value when the ball was at the top of the hill.) Because the [tex]\rm GPE[/tex] of this ball is proportional to its height, at halfway down the hill, the [tex]\rm GPE\![/tex] of this ball would also be [tex](1/2)\![/tex] its initial value.
However, if there was no friction on this ball (and that the ball did not deshape,) that [tex](1/2)[/tex] of the initial [tex]\rm GPE\![/tex] of this ball was not lost. Rather, these [tex](1/2)\![/tex] of the initial [tex]\rm GPE[/tex] would have been converted to the kinetic energy ([tex]\rm KE[/tex]) of this ball.
Hence, when the ball is halfway down the hill:
[tex]\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}[/tex].
[tex]\begin{aligned}& \text{KE halfway down the hill}\\ &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}[/tex].
Therefore:
[tex]\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}[/tex].
In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.
A 45 ml sample of nitrogen gas is cooled from 135 degrees celsius to 15 degrees Celsius in a container that can contract or expand at constant pressure. What is the new volume of nitrogen gas?
Answer:
32 mL
Explanation:
To answer this question we'll use Charles' law, which relates the temperature and volume of a gas at constant pressure:
V₁T₂=V₂T₁
Now we convert the given temperatures to K:
T₁ = 135 °C ⇒ 135 + 273.16 = 408.16 KT₂ = 15 °C ⇒ 15 + 273.16 = 288.16 KAnd given that V₁ = 45 mL, we can input the data:
45 mL * 288.16 K = V₂ * 408.16 KAnd solve for V₂:
V₂ = 32 mLA student is given an antacid tablet that weighs 5.4630 g. The tablet is crushed and 4.3620 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 13.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.7 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the stomach acid (in mL) has been neutralized in the 25.00 mL sample that was titrated
Solution :
It is given that :
Weight of the antacid tablet = 5.4630 g
4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.
25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.
27.7 mL of [tex]$NaOH$[/tex] solution is equivalent to [tex]$25 \ mL$[/tex] of the original stomach acid. Therefore, 13.6 mL of NaOH will take x [tex]$\frac{25\text{ mL of original stomach acid}}{\text{27.7 mL of NaOH}}$[/tex]
= 12.27 ml of the original stomach acid.
1 = 2 Amps, R = 3 Ohms, V = ? Volts
Answer:
6 volts
Explanation:
Use Ohm's Law:
V = IR
V = (2 amps)(3 ohms) = 6 volts
By the reaction between
H3PO4 and Ca(OH)2,
Can be formede:
a. Ca(HPo4)2,
b. СaHP4
c. Ca(H2PO4)2
d. Ca2HP02
e.Сa3(PO4)2
the correct answers : b, c, e
the question is:
I am preparing for an admittion test in chemistry, I face this kind of questions which I dont how to understand or if I should only memorize.
I appreciate all your help.
thank you alot.
Answer:
ANSWER is E
Explanation:
2H₃PO₄ + 3Ca(OH)₂→ 6H₂O + Ca₃(PO₄)₂
Answer:
First step is to find what the products would be. Since this seems to be a double displacement, you get: H3PO4(aq)+Ca(OH)2(aq)→ H2O(L)+Ca3(PO4)2(s)
You know the products because H has a charge of +1, so PO4 must have a charge of -3, and since OH has a charge of -1, Ca must be +2. Then make each compound have a net charge of 0. H20 is a liquid since aq solutions have liquid water. Ca3(PO4)2 is a solid, use the solubility rules.
Lastly, balance the equation: 2H3PO4(aq)+3Ca(OH)2(aq)→ 6H2O(L)+Ca3(PO4)2(s)
So the salt formed is Ca3(PO4)2
how many grams of P3O9 contain 9.67x10^23 atoms of oxygen?
Answer:
49.1 g
Explanation:
First we convert 9.67x10²³ atoms of oxygen into moles, using Avogadro's number:
9.67x10²³ atoms ÷ 6.023x10²³ atoms/mol = 1.60 mol OThen we calculate how many P₃O₉ moles there would be with 1.60 O moles:
1.60 mol O * [tex]\frac{1molP_3O_9}{9molO}[/tex] = 0.18 mol P₃O₉Finally we convert 0.18 P₃O₉ moles into grams, using its molar mass:
0.18 mol P₃O₉ * 237 g/mol = 49.1 gA 9.56 sample of CuS is found to contain 6.35 g of copper what is the percent by mass of the copy and sulfur in this compound? (show your work)
Answer:
%Cu = 66.4%
%S = 33.6%
Explanation:
Step 1: Given data
Mass of CuS: 9.56 gMass of Cu: 6.35 gStep 2: Calculate the mass of S
The mass of CuS is equal to the sum of the masses of Cu and S.
mCuS = mCu + mS
mS = mCuS - mCu
mS = 9.56 g - 6.35 g = 3.21 g
Step 3: Calculate the percent by mass of each element
We will use the following expression.
%Element = mElement / mCompound × 100%
%Cu = mCu / mCuS × 100% = 6.35 g / 9.56 g × 100% = 66.4%
%S = mS / mCuS × 100% = 3.21 g / 9.56 g × 100% = 33.6%
What pair of properties do all solids have?
Answer:
Has definite shape and volume
Solids have high melting point
Explanation:
Which of the following components does a disc brake system use?
brake shoe
adjuster screw
wheel cylinder
caliper
The caliper is a component which is used by a disc brake system for performing its function.
What are the components of disc brake system?The disc braking system consist of components such as disc/rotor, a brake calliper and brake pads. When the brake pedal is pushed, brake fluid creates pressure that produces friction.
So we can conclude that the caliper is a component which is used by a disc brake system for performing its function.
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How many moles are contained in 325 gram sample of ammonium dichromate (NH4)2Cr2O7
Answer:
Hyba odpowiedzi c. Hyba
Mole can be defined as the number of molar mass units of the compound in the sample. The moles of ammonium dichromate in 325 grams sample is 1.29 mol.
What is molar mass?The molar mass can be given as the mass of each atom in the formula unit. The molar mass of ammonium chromate is given as:
[tex]\rm (NH_4)_2Cr_2O_7=2(N)+4\;\times\;2(H)+2(Cr)+7(O)[/tex]
Substituting the mass of each atom to identify the molar mass of ammonium chromate:
[tex]\rm (NH_4)_2Cr_2O_7=2(14)+4\;\times\;2(1)+2(52)+7(16)\\ (NH_4)_2Cr_2O_7=28+8+104+112\\ (NH_4)_2Cr_2O_7=252 \;g/mol[/tex]
The molar mass of ammonium dichromate s 252 g/mol. The moles of the compound in 325 g is given as:
[tex]\rm Moles=\dfrac{mass}{molar\;mass} \\\\Moles=\dfrac{325\;g}{252\;g/mol}\\\\ Moles=1.29\;mol[/tex]
The moles of ammonium dichromate in the sample are 1.29 mol. Thus, option c is correct.
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what is the total of the amount of energy that is reflected back into space
3.
Question
Many chemical warfare reagents readily convert into reactive 3-membered ring intermediates (although
maybe not the Joker's venom from the Batman comics). In turn, these intermediates can damage DNA by
what type of chemical reaction?
Your answer
O a. Elimination
O b. Sn1
OC. SN2
O d. None of the above
Answer:
I think it's letter B
Explanation:
Hope it helps correct if I'm wrong