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 g
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
what is the total of the amount of energy that is reflected back into space
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
how many lone pairs are on the central atom in GeH4?
electronic configuration [Ne]3s2 3p6
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
Which of the following answers is true for the following statement?
The conjugate acid of a strong base is:
O aweak acid
O an amphiprotic acid
O a polyprotic acid
O a lousy (or very weak) acid
O a strong acid
it is either "aweak acid or a lousy (or very weak) acid"
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
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.
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!
I need help with chemistry :(
Answer:
How may we help kind sir
Explanation:
and if this was for points thanks
A 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.
What volume, in mL, of 0.100 M NaOH is required to neutralize 35.0 mL of 0.102 M hydroiodic acid HI?
Answer:
35.7 mL NaOH
Explanation:
M1V1 = M2V2
M1 = 0.100 M NaOH
V1 = ?
M2 = 0.102 M HI
V2 = 35.0 mL
Solve for V1 --> V1 = M2V2/M1
V1 = (0.102 M)(35.0 mL) / (0.100 M) = 35.7 mL NaOH
3) How many grams of calcium hydroxide are necessary to create 49.5 mL of a 1.55 M solution?
Answer:
5.68g
Explanation:
In what type of reaction is water always a product
when hydrogen and oxygen combine then they form water as product
also in neutralization reaction water form as product
hope it helps
Water is always a product in a type of chemical reaction called a neutralization reaction.
A reaction is a process that involves the transformation of one or more substances into one or more different substances.
Neutralization reaction is type of reaction occurs when an acid and a base react together to form a salt and water. The general equation for a neutralization reaction is:
[tex]\rm acid + base \rightarrow salt + water[/tex]
In this type of reaction, the hydrogen ions ([tex]\rm H^+[/tex]) from the acid combine with the hydroxide ions ([tex]\rm OH^-[/tex]) from the base to form water ([tex]\rm H_2O[/tex]). The remaining ions from the acid and the base combine to form a salt.
Therefore, water is always a product in a neutralization reaction, which occurs when an acid and a base react together to form a salt and water.
Learn more about reaction here:
https://brainly.com/question/16737295
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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
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 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.
Learn more about moles, here:
https://brainly.com/question/15209553
A 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%
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 mL6. The drawing shows Samir riding his mountain bike.
i) Draw a circle around the places on the drawing where there
should be a lot of friction.
ii) Explain why there should be a lot of friction in these places.
Answer:
You Need To Upload The Photo Then I Will Help You :)
Explanation:
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
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
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.
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.
What pair of properties do all solids have?
Answer:
Has definite shape and volume
Solids have high melting point
Explanation:
Most cooking utensils are made up of aluminum because aluminum is _____
Explanation:
Cooking utensils, such as pots, pans and menu trays, are often made from aluminium because it is lightweight and conducts heat well, making it energy-efficient for heating and cooling. These properties also make it a preferred material for packaging.
calculate the percentage of sodium and carbon and oxygen in sodium carbonate
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.