Answer: Formula: Mass = (Volume)(Density)
Iron Density = 7.87 g/cm^3
Volume of Iron = 55.2 cm^3
Mass=(V)(D)
Mass= (55.2 cm^3) x (7.87 g/cm^3)
Mass= 434,42 g
Explanation:
Iron has a density of 7.87 g/cm³. 434,42 g is the mass of 55.2 cm³ of iron.
What do you mean by density ?The term density is defined as the measurement of how closely a material is packed together.
It is also defined as the mass per unit volume. Density Symbol is D or ρ Density Formula is ρ = m/V, where ρ is the density, m is the mass of the object and V is the volume of the object.
Density is an important because it allows us to find out what substances will float and what substances will sink when placed in a liquid.
Formula:
Mass = (Volume)(Density)
Given:
Iron Density = 7.87 g/cm³
Volume of Iron = 55.2 cm³
Mass=(V)(D)
Mass= (55.2 cm³) x (7.87 g/cm³)
Mass= 434,42 g
Thus, Iron has a density of 7.87 g/cm³. 434,42 g is the mass of 55.2 cm³ of iron.
To learn more about the density, follow the link;
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What will be the pH of a buffer solution containing an acid of pKa7.5, with an acid concentration exactly one fourth of that of the conjugate base
Answer: pH of buffer solution is 8.1
Explanation:
The formula for the Henderson–Hasselbalch equation is:
[tex]pH=pK_a+\log\frac{[A^-]}{[HA]}[/tex]
[tex]pH[/tex] is the concentration of [tex][H^+][/tex]
[tex]pK_a[/tex] is the acid dissociation constant,
[tex]A^-[/tex] and [tex]HA[/tex] are concentrations of the conjugate base and starting acid.
Putting in the values we get:
[tex]pH=7.5+\log\frac{x}{\frac{x}{4}}[/tex]
[tex]pH=8.1[/tex]
Thus pH of buffer solution is 8.1
The formula for chromium (vi)bromate
Answer: The formula of chromium fluoride is CFr2.
What is the mass in grams of 3.40 x 10 24 atoms he
Answer:
Avogadro's constant says that
1
mole of any atom contains
6.022
⋅
10
23
atoms. In this case you have
3.40
⋅
10
22
atoms:
3.40
⋅
10
22
atoms
6.022
⋅
10
23
atoms
mol
=
5.65
⋅
10
−
2
m
o
l
Step 2
The atomic mass of helium (He) will give you the weight of one mole of this molecule:
1
mol =
4.00
gram:
4.00
g
mol
⋅
5.65
⋅
10
−
2
mol
=
0.226
g
So the
3.40
⋅
10
22
helium atoms weigh
0.226
gram
.
Explanation:
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Write a formula and balanced equation for the following: solid copper metal reacts with aqueous silver nitrate to produce solid silver metal and aqueous copper Nitrate. How do I know if it will form copper (I) nitrate CuNO3 or copper (II) nitrate Cu(NO3)2 ?
Explanation:
copper (II) nitrate
Illustrate variety of substances of which an element can be part:
metal --> blue solution --> blue solid --> black solid --> blue solution (again) --> metal (again).When solid copper reacts with aqueous silver nitrate, the products are copper (II) nitrate and solid silver. and carbon monoxide gas produces solid iron and carbon dioxide gas.
8. 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M
solution. How many mL of 0.800 M can you make?
9. A stock solution of 1.00 M NaCl is available. How many milliliters are needed
to make 100.0 mL of 0.750 M
10.2.00 L of 0.800 M NaNO, must be prepared from a solution known to be 1.50
M in concentration. How many mL are required?
PLS answer ASAP!! WILL MARK AS BRAINLYIST!
In what industry do fertilizers and pesticides wash off and contaminate water supplies?Construction Oil Transportation Agriculture
Answer:
The answer is agriculture.
Explanation:
Answer: Agriculture
Explanation:
I got it right on my exam
Good Morning I have a question I need help and can not find the answer o it maybe someone help me? The question is _______ Most Of the Energy that drives water cycle comes from__________? (this is from Science)
Answer:
hydro, water
Explanation:
Question 7 point
What is the name of the covalent compound with the formula P2N3?
N2 + H2 --> NH3 (unbalanced) How many moles of NH3 are produced when 6.3 moles of H2 gas react with N2 gas?
Answer:
4.2 mol NH3
Explanation:
First, balance your reaction.
N2 + 3H2 --> 2NH3
Multiply 6.3 mol H2 by the mole ratio of 2 mol NH3 for every 3 mol H2 to get moles of NH3 produced.
6.3 mol H2 • (2 mol NH3 / 3 mol H2) = 4.2 mol NH3
what type of reaction occur in cyclohexene
Cyclohexene is a cyclic, six-membered hydrocarbon that contains one double bond. The types of reactions that can occur in cyclohexene would be those that are typical with alkenes generally.
The pi-bonded electrons in the double bond are nucleophilic. So, electrophilic addition reactions could occur with cyclohexene. For example,
cyclohexene + HBr → bromocyclohexane
cyclohexene + H2O/H+ → cyclohexanol
cyclohexene + Br2 → trans-1,2-dibromocyclohexane (racemic)
The latter is a common test for alkenes where one adds bromine to a sample to see if there is decolorization, which would indicate the presence of nucleophilic pi bonds. Bromine, which is dark reddish-brown, will become clear as it reacts with an alkene to form a colorless haloalkane.
Cyclohexene can also be converted to the fully saturated cyclohexane by hydrogenation: cyclohexene + H2/Pd → cyclohexane.
Consider a solution which is 0.10M in NH3 and 0.20M in NH4Cl. Choose ALL of the following that are false. If a small amount of NaOH is added, the pH becomes slightly more basic. If HCl is added, the H ions react with the NH3. If a small amount of HCl is added, the pH becomes slightly more basic. If NaOH is added, the OH- ions react with the NH3. If NH4NO3 is added, the pH becomes more basic.
Answer:
The incorrect or the false options are -
(B) - If NH4NO3 is added, the pH becomes more basic.
(D) - If a small amount of HCl is added, the pH becomes slightly more basic.
(E) - If NaOH is added, the OH- ions react with the [tex]NH_3[/tex].
Explanation:
The given solution the buffer basic .
When a small amount of either is added, the hydrogen ion reacts with the ammonium ion, raising the pH slightly.When a small amount of HCl is added, the hydronium ion reacts with the ammonia, raising the pH slightly.Therefore , the options A ( If a small amount of NaOH is added, the pH becomes slightly more basic.) and C ( If HCl is added, the H+ ions react with the NH3.) All other B , D , E are false .
What mass of nitrogen (N2) is needed to produce 0.125 mol of ammonia (NH3)? Input a numerical answer only. N2 + 3H2 ---> 2NH:
Please help me
A 90-m3 basement in a residence is found to be contaminated with radon coming from the ground through the floor drains. The concentration of radon in the room is 1.5 Bq/L under steady-state conditions. The room behaves as a CSTR and the decay of radon is a first-order reaction with a decay rate constant of 2.09 3 1026 s21 . If the source of radon is closed off and the room is vented with radon-free air at a rate of 0.14 m3 /s, how long will it take to lower the radon concentration to an acceptable level of 0.15 Bq/L
Answer:
25 minutes
Explanation:
V = Volume = [tex]90\ \text{m}^3[/tex]
[tex]C_0[/tex] = Radon concentration under steady state = 1.5 Bq/L
k = Radon decay rate = [tex]2.09\times 10^{-6}\ \text{s}^{-1}[/tex]
[tex]Q[/tex] = Venting rate = [tex]0.14\ \text{m}^3/\text{s}[/tex]
[tex]C_f[/tex] = Final concentration of radon = 0.15 Bq/L
Theoretical detention time is given by
[tex]\theta=\dfrac{V}{Q}\\\Rightarrow \theta=\dfrac{90}{0.14}\\\Rightarrow \theta=642.86\ \text{s}[/tex]
We have the relation
[tex]C_f=C_0e^{-(\dfrac{1}{\theta}+k)t}\\\Rightarrow t=\dfrac{\ln\dfrac{C_f}{C_0}}{-(\dfrac{1}{\theta}+k)}\\\Rightarrow t=\dfrac{\ln\dfrac{0.15}{1.5}}{-(\dfrac{1}{642.86}+2.09\times 10^{-6})}\\\Rightarrow t=1478.25\ \text{s}=\dfrac{1478.25}{60}=24.63\approx 25\text{minutes}[/tex]
The time taken to reach the acceptable level of concentration is 25 minutes.
Suppose you have a 1:1:1 by weight mixture of three solid compounds, salicylic acid 4-nitroaniline naphthalene. You dissolve 1 gram of this mixture in diethyl ether, and place a tiny drop of the ether solution on a TLC plate. After developing the TLC plate, you see three spots. Which compound would would you expect to have the largest Rf value
Answer:
The correct answer is - 4-nitroaniline.
Explanation:
It is given that all three solid compounds salicylic acid + 4-nitroaniline + naphthalene are equal in the ratio in the mixture and then 1 gram of this mixture is dissolved in the diethyl ether and run a drop of the solution on TLC plate. This plate shows three spots.
The salicylic acid and naphthalene would stay dissolved in the diethyl ether solution due to the 4-nitroaniline could be extracted by adding aqueous acid and involve in the aqueous layer and thus spot of 4-nitroaniline would be with largest Rf value.
A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.What is the specific heat of the substance?
Answer:
0.3832
Explanation:
Givens
m = 437.2 grams
c = ?
delta t = 69.8 - 19.3
H = 8460 J
Formula
H = m*c*Δt
Solution
8460 = 437.2 * c * (69.8 - 19.3)
8460 = 437.2 * c * 50.5
8460 = 22078.6 * c
c = 8460 / 22078,6
c = .3832 J / (oC * gr)
when 70.0 grams of mno2 reacted with 128.0 grams of hcl, the reaction resulted in a 62.7% yield of chlorine gas. what is the actual yield of chlorine gas in grams? Mno2 + HCI —> MnCl2 + h2o + cl2
Answer:
35.8g of Cl₂ is the yield
Explanation:
Based on the reaction:
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
1 mole of MnO₂ and 4 moles of HCl react producing 1 mole of Cl₂
To solve this question we must find limiting reactant. with limiting reactant we can find the theoretical yield of Cl₂. As the actual yield is the 62.7% we can find actual yield of Cl₂ in grams:
Moles MnO₂ -Molar mass: 86.9368g/mol-:
70.0g * (1mol / 86.9368g) = 0.805 moles
Moles HCl -Molar mass: 36.46g/mol-:
128.0g * (1mol / 36.46g) = 3.51 moles
For a complete reaction of 3.51 moles of HCl are required:
3.51 moles HCl * (1mol MnO₂ / 4mol HCl) = 0.878 moles MnO₂.
As there are just 0.805 moles of MnO₂, MnO₂ is limiting reactant.
1 mole of MnO₂ produce 1 mole of Cl₂. The theoretical moles of Cl₂ produced are 0.805 moles.
As the yield is of 62.7%, the yield of Cl₂ is:
0.805 moles * (62.7 / 100) = 0.505 moles Cl₂. In grams:
0.505 moles Cl₂ * (70.906g / mol) =
35.8g of Cl₂ is the yieldWhy are some resources, like fossils, found in some places and not in others?
Answer:
Because thats where the animal passed away
Explanation:
A solution contains 1.817 mg of CoSO4 (155.0 grams/mole) per mL. Calculate the volume (in mL) of 0.009795 M Zn2 needed to titrate the excess complexing reagent after the addition of 70.00 mL of 0.009005 M EDTA to a 20.00 mL aliquot of the Co2 solution.
Answer:
85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .
Explanation:
Lets calculate
After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml
As the number of moles of [tex]CoSO_4[/tex] =[tex]\frac{Given mass }{molar mass}[/tex]
=[tex]\frac{1.817}{155}[/tex]
=0.01172
Molarity = [tex]\frac{no. of moles}{volume of solution}[/tex]
=[tex]\frac{0.01172}{20}[/tex]
=0.000586 moles
Excess of EDTA = concentration of EDTA - concentration of CoSO4
= 0.009005 - 0.000586
= 0.008419 M
As M1V1 ( Excess of EDTA ) = M2V2 [tex](Zn^2^+)[/tex]
[tex]0.008419\times100ml=0.009795\times V2[/tex]
[tex]V2=\frac{0.008419\times100}{0.009795}[/tex]
V2 =85.952 ml
Therefore , 85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .
A substance that contains two or more kinds of atoms is
HELP ASAP PLEASE!!! :)
Answer:
Alkane
Explanation:
Alkenes have CH in their condensed formulas and alkynes have C in their condensed formulas. There are only CH3 and CH2 so it's an alkane.
6. (5 pts) From among the isomeric compounds with molecular formula C4H9Cl, identify the one having a 1H NMR spectrum that: a) contains only a single peak b) has several peaks, including a doublet at d 3.4 c) has several peaks, including a triplet at d 3.5 d) Has several peaks, including two distinct signals, each with an integration of 3H, one of them a triplet at d 1.0 and the other a doublet at d 1.5.
Answer:
Explanation:
Chemically identical protons in 1HNMR spectra are protons that are found in the same chemical system and produce one signal in the spectra. Non-equivalent protons are protons that reside in various chemical environments and display different signals.
(a) The provided alkyl halide has four carbons, nine hydrogens, and one chlorine atom, according to the molecular formula. To achieve a single NMR peak, all protons (9 hydrogens) must be in the same chemical environment. That is, the carbon skeleton should be made up of three methyl groups, each bound to a single carbon atom.
As a consequence, the tert-butyl chloride structure can be present. That it has three methyl groups that are all the same and are all bound to the same carbon atom. As a result. A single peak is generated by proton NMR.
(b) The other isomeric variant of C4H9Cl, which has several peaks and a doublet at d3.4, could have a proton at a nearby carbon. And then does it produce a doublet. To create the doublet, two methyl groups should be substituted for the carbon. The value of 3.4 indicates that the protons can be exposed to chlorine. As a result, the compound's composition is iso-butyl chloride.
(c) The compound C4H9Cl can have two protons on adjacent carbons since it has multiple peaks and a triplet at d3,5. And then does it generate a triplet. The meaning of 3.5 indicates that the protons should be exposed to chlorine. As a result, the compound's composition is n-butyl chloride.
(d) The compound C4H9Cl has many peaks, including a methyl group triplet at d1.0 and a methyl group doublet after. It's in the same family as the -CH party. Due to the chlorine atom, this doublet occurs at downfield at 1.5. As a result, the compound's composition is sec-butyl chloride.
Nitric oxide (NO) reacts with oxygen gas to produce nitrogen dioxide. A gaseous mixture contains 0.66 g of nitric oxide and 0.58 g of oxygen gas. After the reaction is complete, what mass of nitrogen dioxide is formed? Which reactant is in excess? How do you know? Suppose you actually recovered 0.91 g of nitrogen dioxide. What is your percent yield?
Answer:
NO is the limiting reagent.
In this reaction 0.886 mole of NO2 is produced
Explanation:
The chemical equation for this reaction is
2NO(g) + O2(g) → 2NO2(g)
In this limiting reagent reaction, 2 moles of NO reacts with one mole of O2 to produce 2 mole of 2NO2
0.886 mole of NO * (2 mole of NO2/2 mole of NO) = 0.886 mole of NO2
0.503 mole of O2 * (1 mole of NO2/1 mole of O2) = 1.01 mole of NO2
Hence, NO is the limiting reagent.
In this reaction 0.886 mole of NO2 is produced
how many molecules of sodium chloride are in 2.5 moles??
Answer:
1.5x1024
Explanation:
Data Collection
Mass of the original sample of mixture (g) 1.558
Mass of recovered naphthalene (g) 0.483
Mass of recovered 3-nitroaniline (g) 0.499
Mass of recovered benzoic acid (g) 0.467
Calculations:
a. % by mass of naphthalene in original sample.
b. % by mass of 3-nitroaniline in original sample.
c. % by mass of benzoic acid in original sample.
d. total percent recovered.
Answer:
For a): The mass percent of naphthalene in the original sample is 31.00 %.
For b): The mass percent of 3-nitroaniline in the original sample is 32.03 %.
For c): The mass percent of benzoic acid in the original sample is 29.97 %.
For d): The total percent recovered is 93.00 %.
Explanation:
Percentage by mass is defined as the ratio of the mass of a substance to the mass of the solution multiplied by 100. The formula used for this is:
[tex]\text{Percent by mass}=\frac{\text{Mass of substance}}{\text{Mass of a solution}} \times 100[/tex] ......(1)
a):
Mass of naphthalene = 0.483 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\%\text{ mass of naphthalene}=\frac{0.483 g}{1.558}\times 100\\\\\%\text{ mass of naphthalene}=31.00 \%[/tex]
b):
Mass of 3-nitroaniline = 0.499 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\%\text{ mass of 3-nitroaniline}=\frac{0.499 g}{1.558}\times 100\\\\\%\text{ mass of 3-nitroaniline}=32.03 \%[/tex]
c):
Mass of benzoic acid = 0.467 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\%\text{ mass of benzoic acid}=\frac{0.467 g}{1.558}\times 100\\\\\%\text{ mass of benozic acid}=29.97 \%[/tex]
d):
Total mass recovered = [0.483 + 0.499 + 0.467] = 1.449 g
Mass of the sample = 1.558 g
Plugging values in equation 1:
[tex]\text{Total percent recovered}=\frac{1.449 g}{1.558}\times 100\\\\\text{Total percent recovered}=93.00\%[/tex]
Research and describe one career in health and fitness that you would consider
Answer:
a PE teacher would be in both catagories of heath and fitness
Explanation:
How many moles of sodium sulfate are produced along with 5.86 moles of iron(III)
hydroxide?
Fe2(SO4)3 + 6NaOH → 3Na SO4 + 2Fe(OH)3
Answer:
5.86mol 2Fe(OH)3 * [3 mols Na SO4 / 2 mols Fe(OH)3] = 8.79 mols sodium sulfate
Explanation:
To solve this question you need to look at the ratio of sodium sulfate to iron(III) hydroxide. The ratio is 3 to 2, meaning that for every 3 moles of Na SO4 there are also 2 moles of Fe(OH)3. So, you can multiply 5.86 by (3/2) to find the mols of sodium sulfate.
A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by
mass.
Question 1: The empirical formula for this compound is :
Question 2: The molar mass for this compound is 82.98 g/mol.
The Molecular formula for this compound is:
Answer: 1. The empirical formula is [tex]PNF_2[/tex]
2. The molecular formula is [tex]PNF_2[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of P = 37.32 g
Mass of N = 16.88 g
Mass of F = 45.79 g
Step 1 : convert given masses into moles.
Moles of P =[tex]\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles[/tex]
Moles of F =[tex]\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For P = [tex]\frac{1.20}{1.20}=1[/tex]
For N = [tex]\frac{1.20}{1.20}=1[/tex]
For F =[tex]\frac{2.41}{1.20}=2[/tex]
The ratio of P: N: F= 1: 1: 2
Hence the empirical formula is [tex]PNF_2[/tex]
The empirical weight of [tex]PNF_2[/tex]= 1(31)+1(14)+2(19)= 82.98 g.
The molecular weight = 82.98 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1[/tex]
The molecular formula will be=[tex]1\times PNF_2=PNF_2[/tex]
Which of the following must be overcome by a rocket's thrust?
Answer:
the payloads weight
Explanation:
a p e x on the ground stream
Answer:
The payload's weight
Explanation:
i just took the test on a pex :)
Calculate the percent composition of C in carbon dioxide.
Answer:
27.27%
Explanation:
How many moles of hydrogen
are in 3.06 x 10^-3 g of glycine C2H5NO2
Answer:
2.04x10⁻⁴ mol
Explanation:
First we convert 3.06x10⁻³ grams of glycine into moles of glycine, using its molar mass:
3.06x10⁻³ g ÷ 75 g/mol = 4.08x10⁻⁵ mol C₂H₅NO₂In order to calculate the number of hydrogen moles, we multiply the number of glycine moles by 5, as there are 5 hydrogen moles per glycine mol:
4.08x10⁻⁵ mol C₂H₅NO₂ * 5 = 2.04x10⁻⁴ mol H