Answer:
(c) Nitric acid [tex]\rm HNO_3\, (aq)[/tex] is the limiting reactant.
(d) Approximately [tex]6.6\; \rm g[/tex] of [tex]\rm CuO\, (s)[/tex] will be in excess.
(e) The percentage yield of [tex]\rm Cu(NO_3)_2[/tex] is approximately [tex]91\%[/tex]. (Rounded to two significant figures, as in other quantities in the question.)
Explanation:
Start with the balanced chemical equation:
[tex]\rm CuO\, (s) + 2\; HNO_3\, (aq) \to Cu(NO_3)_2\, (aq) + H_2O\, (l)[/tex].
Look up relevant relative atomic mass data on a modern periodic table:
[tex]\rm Cu[/tex]: [tex]63.546[/tex].[tex]\rm O[/tex]: [tex]\rm 15.999[/tex].[tex]\rm H[/tex]: [tex]1.008[/tex].[tex]\rm N[/tex]: [tex]14.007[/tex].Calculate the formula mass of the species:
[tex]M(\mathrm{CuO}) = 63.546 + 15.999 = 79.545\; \rm g\cdot mol^{-1}[/tex].[tex]M(\mathrm{Cu(NO_3)_2}) = 63.546 + 2\times (14.007 + 3 \times 15.999) = 187.554\; \rm g\cdot mol^{-1}[/tex].Limiting ReactantThere are two reactants in this reaction: [tex]\rm CuO[/tex] and [tex]\rm HNO_3\, (aq)[/tex]. Assume that [tex]\rm CuO\![/tex] is the limiting one. In other words, assume that all the [tex]\rm CuO\!\![/tex] is consumed before [tex]\rm HNO_3\, (aq)\![/tex] was.
Consider: how many moles of [tex]\rm HNO_3\, (aq)\!\![/tex] would be required to convert all that [tex]7.0\; \rm g[/tex] of [tex]\rm CuO\!\!\![/tex] to [tex]\rm Cu(NO_3)_2[/tex]?
Calculate the number of moles of [tex]\rm CuO[/tex] formula units in [tex]7.0\;\rm g[/tex] of
[tex]\displaystyle n({\rm CuO}) = \frac{m}{M} = \frac{7.0\; \rm g}{79.545\; \rm g \cdot mol^{-1}} \approx 0.088001\; \rm mol[/tex].
Note the ratio between the coefficients of [tex]\rm CuO\, (s)[/tex] and [tex]\rm HNO_3\, (aq)[/tex]:
[tex]\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{CuO\, (s)})} = \frac{2}{1} = 2[/tex].
Therefore:
[tex]\begin{aligned}&n(\text{$\mathrm{HNO_3\, (aq)}$, consumed (under assumption)})\\ &= \frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\cdot n(\mathrm{CuO\, (s)}) \\ &\approx 2 \times 0.088001\; \rm mol \approx 0.17600\; \rm mol\end{aligned}[/tex].
On the other hand, how many moles of [tex]\rm HNO_3\, (aq)[/tex] are actually available?
Convert the volume of that [tex]\rm HNO_3\, (aq)[/tex] solution to the standard unit (liter.)
[tex]V(\rm HNO_3\, (aq)) = 50\; \rm mL = 0.050\; \rm L[/tex].
Calculate the number of moles of [tex]\rm HNO_3\, (aq)[/tex] in that [tex]0.20\; \rm M[/tex] solution:
[tex]n({\rm HNO_3\, (aq)}) = c\cdot V = 0.050\; \rm L \times 0.20\; \rm mol \cdot L^{-1} = 0.010\; \rm mol[/tex].
Apparently, the quantity of [tex]\rm HNO_3\, (aq)[/tex] required exceeded the quantity that is available. Therefore, the assumption is invalid, and [tex]\rm CuO[/tex] cannot be the limiting reactant. At the same time,
Mass of the reactant in excessSince it is now known that all that [tex]0.010\; \rm mol[/tex] of [tex]\rm HNO_3\, (aq)[/tex] will be consumed, apply that coefficient ratio again to obtain the quantity of [tex]\rm CuO[/tex] consumed in this reaction:
[tex]\begin{aligned}&n(\text{$\mathrm{CuO}$, consumed})\\ &= n(\text{$\mathrm{HNO_3\, (aq)}$, consumed}) \left/\frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\right. \\ &\approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol\end{aligned}[/tex].
It was already shown that the formula mass of [tex]\rm CuO[/tex] is (approximately) [tex]79.545\; \rm g \cdot mol^{-1}[/tex]. Therefore, the mass of that [tex]0.0050\; \rm mol[/tex] formula units of [tex]\rm CuO\![/tex] would be:
[tex]\begin{aligned}& m(\text{$\rm CuO$, consumed}) \\&= n \cdot M \approx 0.0050\; \rm mol \times 79.545\; \rm g\cdot mol^{-1} \\&\approx 0.39773\; \rm g\end{aligned}[/tex].
Before the reaction, [tex]7.0\; \rm g[/tex] of [tex]\rm CuO[/tex] is available. Therefore, all that [tex]7.0\; \rm g - 0.39773\; \rm g \approx 6.6\; \rm g[/tex] of [tex]\rm CuO\![/tex] would be in excess.
Percentage YieldSimilarly:
[tex]\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{Cu(NO_3)_2\, (aq}))} = \frac{2}{1} = 2[/tex].
Apply this ratio to find the theoretical yield of [tex]\rm Cu(NO_3)_2[/tex]:
[tex]n(\text{$\mathrm{Cu(NO_3)_2\, (aq)}$, theoretical yield}) \approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol[/tex].
Find the mass of that [tex]0.0050\; \rm mol[/tex] of [tex]\rm Cu(NO_3)_2[/tex] formula units using its formula mass:
[tex]m(\text{$\rm Cu(NO_3)_2\, (aq)$, theoretical yield}) \approx 0.93777\; \rm g[/tex].
Calculate the percentage yield given that the actual yield is [tex]0.85\; \rm g[/tex]:
[tex]\begin{aligned}&\text{Percentage Yield} \\ &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\% \\ &= \frac{0.85\; \rm g}{0.93777\; \rm g} \times 100\% \approx 0.91\% \end{aligned}[/tex].
(Rounded to two significant figures.)
(c) The limiting reactant is HNO₃
(d) The mass of the excess reactant after the reaction is approximately 6.6 grams
(e) The percentage yield of copper (II) nitrate from the reaction is approximately 90.64%
The reason the above values are the correct values is as follows;
The given parameters are;
The mass of copper(II)oxide in the reaction = 7.0 g
The volume of the 0.20 M nitric acid, HNO₃ = 50 mL
(c) Concentration of the reactants
The molar mass of CuO = 79.545 g/mol
Number of moles = Mass/(Molar mass)
The number of moles of CuO = (7 g)/79.545 g/mol ≈ 0.088 moles
50 mL of 0.20 M HNO₃, contains 50/1000 × 0.2 = 0.01 moles of HNO₃
The chemical equation for the reaction is CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O
Therefore;
One mole of CuO reacts with two moles of HNO₃ to produce one mole of Cu(NO₃)₂ and one mole of H₂O
Therefore, 0.088 moles of CuO reacts with 2 × 0.088 = 0.176 moles of HNO₃
Given that there is only 0.01 moles of HNO₃, the limiting reactant is the HNO₃, which is not enough to completely react with the CuO which is the excess reactant
(d) The mass of the CuO that reacts with the 0.01 moles of HNO₃ is given as follows;
1 mole of CuO reacts with 2 moles HNO₃
0.01 moles of HNO₃ will react with 0.01/2 = 0.005 moles of CuO
Mass = Number of moles × Molar mass
The mass of 0.005 moles of CuO = 0.005 moles × 79.545 g/mol = 0.397725 grams
The mass of the CuO left = Initial mass - Reacting mass
∴ The mass of the CuO left = 7 grams - 0.397725 grams = 6.602275 grams
The mass of the excess reactant (CuO) after the reaction ≈ 6.6 grams
(e) The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = Half the number of moles of HNO₃ in the reaction
The number of moles of HNO₃ in the reaction = 0.01 moles
∴ The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = (1/2) × 0.01 moles = 0.005 moles
The molar mass of Cu(NO₃)₂ = 187.56 g/mol
The theoretical mass of Cu(NO₃)₂ produced = 0.005 moles × 187.56 g/mol = 0.9378 grams
The actual yield of copper (II) nitrate is 0.84 g
[tex]Percentage \ yield = \mathbf{\dfrac{Actual \ yield}{Theoretical \ yield} \times 100}[/tex]
Therefore;
[tex]\% \ yield \ of \ Cu(NO_3)_2 = \dfrac{0.85 \, g}{0.9378 \, g} \times 100 \approx 90.64 \%[/tex]
The percentage yield of copper (II) nitrate, %yield ≈ 90.64%
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A sample of mass 6.814 grams is added to another sample weighing 0.08753 grams.
The subsequent mixture is then divided into exactly 3 equal parts. One of those parts has the yield multiplied by 7.6335 times; what is the final mass?
Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
classify the organisms based on what they eat. producer consumer decomposer
Lowest - Producer - plants (mostly)
They get food by converting sunlight into nutrients
Middle - Consumer - example animals
They eat other living things such as plants or other animals for their nutrients
Top - Decomposer - bacteria/microorganisms
They decompose dead body
sorry if im wrong
Which pair of compounds will form a buffer in aqueous solution? a) HCN and NaCN b) HCl and NaOH c) NaCN and NaOH d) HCN and HCl e) HCl and NaCl f) NaCN and KCN
Answer:
a) HCN and NaCN
Explanation:
Buffer solution is made by mixing either a weak base and its salt like ammonia and ammonium chloride or weak acid and its salt like HCN and its salt NaCN or acetic acid and sodium acetate .
In the given option , all of the rest of option has either strong acid or strong base or both . In the last option both are salt so it can not form buffer solution .
It is capable of keeping the pH of the solution constant .
Write the numbers in scientific notation.
291.7 = 2.917
X10%
where x =
0.0960 -
X10%
where x =
The numbers in scientific notation are 291.7 = 2.917 × 10₂ and 0.0960 = 9.60 × 10⁻².
Scientific notation is commonly used in scientific and mathematical calculations, as well as when dealing with very large or very small numbers. It allows for a more compact and manageable representation of these numbers.
To write the numbers in scientific notation, we need to express them in the form of "a × 10^b," where "a" is a number between 1 and 10, and "b" is an integer.
For the number 291.7, write it in scientific notation as:
291.7 = 2.917 × 10²
For the number 0.0960, write it in scientific notation as:
0.0960 = 9.60 × 10⁻²
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Based on Figure 25-2, are red giants hotter than white dwarfs?
Answer:
No
Explanation:
A white dwarf has a higher surface temperature than a red giant.
is another way in which nitrogen gets fixed in the soil
Answer:
through Rhizobium bacteria
Adding charts and graphs helps a scientist
O To state the problem
O To determine trends
O To simplify results
O Both B and C
O All of the above
Answer:
b and c
Explanation:
the problem was solved through the experiment and tested
When comparing lithium, sodium, and potassium to calcium, strontium, and barium, early chemists noticed that:
Answer:
Explanation:
Lithium, Potassium and Sodium follow each other in that order when it comes to reactivity.
Succinctly put, Lithium is the most reactive. Potassium is more reactive and Sodium is the least reactive.
Also, for Calcium, Strontium and Barium. It is the least reactive in this order.
Strontium, Barium, Calcium
That is, Strontium is more reactive than Barium, which is more reactive than Calcium. Calcium is the least reactive of the 3
How many mL are in 1.000 cup?
Answer:
236588
Explanation:
1000 x 237 for an approx. amount
Answer:
236.588 mL in 1 cup.
Explanation:
For this problem, you must know that 1 cup is equal to 8 ounces and that 1 ounce is 29.5735 mL. Using this information, let' perform the conversion.
1 cup * ( 8 oz / 1 cup ) * ( 29.5735 mL / 1 oz) == ? mL
( 1 cup * 8 oz * 29.5735 mL ) / ( 1 cup * 1 oz) == ? mL
236.588 mL == ? mL
Hence, there are 236.588 mL in 1 cup.
Cheers.
How does the amount of sunlight affect the number of eggs laid by a chicken?
Explanation:
The amount of daylight hours affects a chicken's reproductive cycle. Hens will begin laying when the amount of daylight reaches 14 hours per day during early spring. Maximum egg laying will occur when the day length reaches 16 hours per day
based on the techniqes yo have learned in the organic chemistry lab how would you seperate any unreated alcohol from the ester
Answer:
Fractional distillation and HP-LC
Explanation:
This is a technique useful for analytes with close boiling points. Any alcohol-ester azotopes can be further refined using high-performance liquid chromatography (HP-LC) column.
The exhaust gas from an automobile contains 3% by volume of carbon monoxide (CO). Express this concentration in mg/m3 at 25oC and 1 atm.
Answer:
[tex]24540\frac{mg}{m^3}[/tex]
Explanation:
Hello,
In this case, since the 3% by volume is represented as:
[tex]\frac{3L\ CO}{L\ gas}[/tex]
By using the ideal gas equation we compute the density of CO:
[tex]\rho =\frac{MP}{RT} =\frac{28g/mol*1atm}{0.082\frac{atm*L}{mol*K}*298K}= 0.818g/L[/tex]
Then we apply the conversion factors as follows:
[tex]=\frac{3L\ CO}{100L\ gas}*\frac{0.818g\ CO}{1L\ CO} *\frac{1000mg\ CO}{1g\ CO} *\frac{1000L\ gas}{1m^3\ gas} \\\\=24540\frac{mg}{m^3}[/tex]
Regards.
Some people argue that a policy of putting out wildfires is having a negative long-term effect on the number and strength of wildfires. Suggest why this could be the case.
Answer:
The reason for this is that putting out the fires only postpones the fire outbreak to a later date, and there is the fear of the fire outbreak being more sever when it actually comes.
When wildfires burn, they clean up the forest off dead trees and falling logs. Also dried leaves and twigs and unnecessarily dense vegetation is cleared up by the fire. These materials are the main fuel of these wildfires. Putting out these fires, especially those that start naturally means that these fuel that should be cleaned up are allowed to accumulate so that when the fire actually happens, it does so with an unnatural intensity. Also, when fire burns, the ashes that are left act as nutrition for the forest, and the forest is allowed to regrow; reborn from the ashes. The only cases that might need human intervention is when the fire is human caused or due to human activities. Natural causes of fire can be due to a very high temperature, lightning striking a tree, etc.
sort the sequences based on the types of mutation they display
Answer:
There are three types of DNA Mutations: base substitutions, deletions and insertions. Single base substitutions are called point mutations, recall the point mutation Glu -----> Val which causes sickle-cell disease. Point mutations are the most common type of mutation and there are two types.
Cell membranes are largely hydrophobic structures. Which compound will pass through a membrane more easily, glucose or 2,4-dinitrophenol? Explain.
Answer:
2,4-dinitrophenol
Explanation:
A hydrophobic molecule does not dissolve nor mix with water. The cell membranes are made up of hydrophobic lipids. Hence, hydrophobic molecules easily pass through the cell membrane.
Glucose is partly hydrophilic hence it is transported across the cell membrane by special proteins. However, 2,4-dinitrophenol is a hydrophobic molecule thus it is easily transported across the cell membrane.
feel free to answer any of these. i need the answer to the appropriate amount of significant figures on this pleasee
Answer:
6. (2.7*[tex]10^{5}[/tex])
8. (1.0*[tex]10^{4}[/tex])
Explanation:
Johnny was finished with his
experiment so he placed all of his
unused chemicals back into their
original containers. What should
Johnny have done instead?
The sono
What could be some disadvantages to this scientific method?
Answer:
Nothing has full knowledge of the world.
Senses can deceive us - science only provides us with an incomplete picture of the world.
Scientists can never be completely unbiased.
Science isn't free from error.
No way of knowing what is real - some things could be illusions
Explanation:
Which of the following correctly describes a mixture? (4 points) a The particles are chemically bonded together, and they retain their individual physical and chemical properties. b The particles are not chemically bonded, and they can only combine in certain set ratios. c The particles do not retain their individual chemical properties, and they can only be separated by chemical means. d The particles have no set ratio for how to combine, and they can be separated by physical means.
Answer:
the answer is B the particles are not chemically bonded, they can only be combined in a certain set ratios
Explanation:
Two atoms of hydrogen combine to form a molecule
of hydrogen gas, the energy of the H2 molecule is :
(1) Greater than that of seperate atoms
(2) Equal to that of seperate atoms
(3) Lower than that of seperate atoms
(4) Some times lower and some times higher
Answer:
lower than the separate atoms. Because an electron in this molecular orbital is strongly attracted to both nuclei, the electron is more stable (at lower energy) than it is in the 1s orbital of the hydrogen atom. Because it concentrates electron density between the nuclei, the bonding molecular orbital holds the atoms together in a covalent bond.
Please Help Me Which ink spot had the most pigments? Question 3 options: A) red B) blue C) black D) green
Answer:
Black
Explanation:
A 1.75 gram perfect cube of wood has a side measure of 2.45 cm, what is the density of the wood?
Answer:
0.119 g/cm³
Explanation:
Step 1: Given data
Mass of the cube (m): 1.75 g
Side measure (l): 2.45 cm
Step 2: Calculate the volume of the cube
We will use the following expression.
V = l³
V = (2.45 cm)³
V = 14.7 cm³
Step 3: Calculate the density (ρ) of the cube
The density of the cube is equal to its mass divided by its volume.
ρ = m/V
ρ = 1.75 g/14.7 cm³
ρ = 0.119 g/cm³
Which statement best describes the types and locations of particles that make up the atom? A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom. B. The neutral-charged neutrons, positive-charged protons, and negative-charged electrons are all found within the nucleus of the atom. C. The negative-charged electrons and positive-charged protons are found within the nucleus, and the neutral-charged neutrons orbit outside the nucleus of the atom. D. The negative-charged neutrons and positive-charged protons are found within the nucleus, and the neutral-charged electrons orbit outside the nucleus of the atom.
Answer:
A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom.
Explanation:
An atom basically consists of three sub atomic particles. They are; Neutrons, Electrons and Protons.
The Electrons are negatively charged and can be found outside the nucleus of the atom,
The Protons are positively charged and can be found in the nucleus of an atom.
The Neutrons are neutral in terms of charged and can be found in the nucleus along with the protons.
The option that is correct is;
A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom.
A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of the solution. Round your answer to decimal places.
Answer:
[tex]pH=4.1[/tex]
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
[tex]pH=pKa+log(\frac{[base]}{[acid]} )[/tex]
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
[tex]pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1[/tex]
Regards.
Calculate the density of an object with a mass of 220 and a volume of 145mL.
Answer:
1.52 g/mL
Explanation:
The formula for finding the density of an object is d = m/v, or density = mass divided by volume.
Therefore, we can input the two values and solve for d.
d = 220/145
d = 44/29
d ≈ 1.52 g/mL
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0 10 20 30 40 conc B/(mol/L) 0 0.089 0.153 0.200 0.230 0.312
1) Determine the order of the reaction. (6 pts.)
2) Find its rate constant. 19 pts.) Note: no unit is needed, just the numerical answer. Hint: convert your minutes to seconds.
Answer:
1) The order of the reaction is of FIRST ORDER
2) Rate constant k = 5.667 × 10 ⁻⁴
Explanation:
From the given information:
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.
liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.
The following data was obtained:
t/min 0 10 20 30 40 ∞
conc B/(mol/L) 0 0.089 0.153 0.200 0.230 0.312
For a first order reaction:
[tex]K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})[/tex]
where :
K = proportionality constant or the rate constant for the specific reaction rate
t = time of reaction
[tex]C_o[/tex] = initial concentration at time t
[tex]C _{\infty}[/tex] = final concentration at time t
[tex]C_t[/tex] = concentration at time t
To start with the value of t when t = 10 mins
[tex]K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})[/tex]
[tex]K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})[/tex]
[tex]K_1 =0.03358 \ min^{-1}[/tex]
[tex]K_1 \simeq 0.034 \ min^{-1}[/tex]
When t = 20
[tex]K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})[/tex]
[tex]K_2= 0.05 \times \ In ( 1.9623)[/tex]
[tex]K_2=0.03371 \ min^{-1}[/tex]
[tex]K_2 \simeq 0.034 \ min^{-1}[/tex]
When t = 30
[tex]K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})[/tex]
[tex]K_3= 0.0333 \times \ In ( \dfrac{0.312}{0.112})[/tex]
[tex]K_3= 0.0333 \times \ 1.0245[/tex]
[tex]K_3 = 0.03412 \ min^{-1}[/tex]
[tex]K_3 = 0.034 \ min^{-1}[/tex]
When t = 40
[tex]K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})[/tex]
[tex]K_4=0.025 \times \ In ( \dfrac{0.312}{0.082})[/tex]
[tex]K_4=0.025 \times \ In ( 3.8048)[/tex]
[tex]K_4=0.03340 \ min^{-1}[/tex]
We can see that at the different time rates, the rate constant of [tex]k_1, k_2, k_3, and k_4[/tex] all have similar constant values
As such :
Rate constant k = 0.034 min⁻¹
Converting it to seconds ; we have :
60 seconds = 1 min
∴
0.034 min⁻¹ =(0.034/60) seconds
= 5.667 × 10 ⁻⁴ seconds
Rate constant k = 5.667 × 10 ⁻⁴
20g of ideal gas contains only atoms of s and o occupies 5.6l at ntp what is the mol wt of gas
Answer:
20g gas contains and this so allow in tha 36G in per laar
Explanation:
A flask that weighs 345.8g is filled with 225mL of carbon tetrachloride. The weight of the flask and carbon tetrachloride is found to be 703.55g. From this information, calculate the density of carbon tetrachloride.
Answer:
Density = 1.59 g/mLExplanation:
The density of carbon tetrachloride can be found by using the formula
[tex]Density = \frac{mass}{volume} [/tex]From the question
Volume = 225 mL
To find the mass of carbon tetrachloride in the flask subtract the mass of the flask from the total mass of the flask and carbon tetrachloride
That's
mass of carbon tetrachloride =
703.55 - 345.8
= 357.75 g
Substitute the values into the above formula and solve for the density
That's
[tex]Density = \frac{357.75}{225} [/tex]We have the final answer as.
Density = 1.59 g/mL
Hope this helps you
Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added to 28.0 mL of 0.230 M NaOH(aq) .
Answer:
pH = 12.35
Explanation:
The HCl reacts with NaOH, thus:
HCl + NaOH → NaCl + H₂O
1 mole of HCl reacts per mole of NaOH
To know the pH we must know the moles of NaOH and the moles of HCl that reacts
Moles NaOH and HCl
NaOH: 0.0280L * (0.230mol / L) = 6.44x10⁻³ moles
HCl: 0.0230L * (0.230mol / L) = 5.29x10⁻³
As moles NaOH > Moles HCl, the moles that remains of NaOH are:
Moles NaOH = 6.44x10⁻³ - 5.29x10⁻³ = 1.15x10⁻³ moles NaOH = Moles OH⁻
In 23.0mL + 28.0mL = 51.0mL = 0.0510L:
1.15x10⁻³ moles OH⁻ / 0.0510L =
0.0225M OH⁻
pOH = -log [OH⁻]
pOH = 1.647
And pH = 14 - pOH
pH = 12.352The higher the numbers on the NFPA diamond, the more _____ the substance is.
ARare incorrect answer
BDangerous incorrect answer
CPure incorrect answer
DSafe
The correct answer is B. Dangerous
Explanation:
The NFPA diamond is used to show how safe or dangerous a substance or material is based on four main categories: Health (blue), Flammability (red), Instability (yellow), special hazards (white). Moreover, the three first hazards are expressed using numbers from 0 to 4, in which 0 means no hazard while 4 is the most dangerous category. For example, in the health square finding 0 means the substance represents no hazard for health, while 4 means the substance is lethal. This means the higher the number in this diamond, the most dangerous the substance is.