Answer:
T₂ = 16.7 °C
Explanation:
Given data:
Initial volume = 150.0 mL
Temperature = 25.0 °C
Final volume = 100 mL
Final temperature = ?
Solution:
V₁ /T₁ = V₂/T₂
T₂ = V₂ . T₁/V₁
T₂ = 100 mL .25 °C / 150.0 mL
T₂ = 2500 mL. °C / 150.0 mL
T₂ = 16.7 °C
Final Answer:
Upon cooling from a volume of 150.0 mL to 100.0 mL at constant pressure, the temperature of the carbon monoxide gas is estimated to be approximately -74.4 °C, as determined using Charles's Law.
Explanation:
To ascertain the new temperature of carbon monoxide as it undergoes cooling from 150.0 mL to 100.0 mL at a constant pressure, Charles's Law is applied. Charles's Law asserts that the volume of a gas is directly proportional to its temperature when pressure remains constant, and this relationship can be expressed mathematically as V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature.
Given an initial temperature (T1) of 25.0 °C, it needs conversion into kelvins by adding 273.15 to the Celsius temperature, resulting in T1 = 25 + 273.15 = 298.15 K. The initial volume (V1) is 150.0 mL, and the final volume (V2) is 100.0 mL. Utilizing Charles's Law, we can now determine the final temperature (T2).
Applying Charles's Law, we derive:
V1/T1 = V2/T2
(150.0 mL) / (298.15 K) = (100.0 mL) / (T2)
T2 = (100.0 mL) * (298.15 K) / (150.0 mL)
T2 = 198.7667 K
To express the temperature in Celsius, we subtract 273.15 from the temperature in kelvins:
T2 = 198.7667 K - 273.15
T2 ≈ -74.3833 °C
Thus, the new temperature of the carbon monoxide after cooling is approximately -74.4 °C.
Which chemical equation represents a precipitation reaction?
Answer:
See the answer below, please.
Explanation:
A precipitation reaction is defined as one that yields an insoluble compound (precipitate) as a product by mixing two different solutions. An example:
NaCl + AgN03 -> Agcl (precipitate) + NaN03
Answer:
K2CO3 + PbCl2 → 2KCl + PbCO3
Explanation:
This is correct because a percipitant solution is one that has an insoluble compound in it and PbCO3 is insoluble since the Pb is not amonium or an alkaline metal. The compound PbCO3 is unsoluable and 2KCl is soluable.
SOLUBILITY RULES:
1.) Hydroxides (OH−), carbonates (CO32−), and phosphates (PO43−) are insoluble, except for compounds containing group 1 alkali metals and ammonium (NH4+).
2.) Chlorides (Cl−), bromides (Br−), and iodides (I−) are soluble, except for compounds containing silver (Ag+), mercury(I) (Hg22+), and lead (Pb2+).
How many decigrams are in 4.6 decagrams?
0.0046 decigrams
0.046 decigrams
460 decigrams
4,600 decigrams
Answer:
[tex]\large \boxed{\text{460 dg}}[/tex]
Explanation:
1. Convert decagrams to grams
[tex]\text{Mass} = \text{4.6 dag} \times\dfrac{\text{10 g}}{\text{1 dag}} = \text{46 g}[/tex]
2. Convert grams to decigrams
\[tex]\text{Mass} = \text{46 g} \times\dfrac{\text{10 g}}{\text{1 dg}} = \text{460 dg}\\\\\large \boxed{\textbf{4.6 dag = 460 dg}}[/tex]
What's C2H6O2 in empirical formula?
The empirical formula for [tex]C_2H_6O_2[/tex] is [tex]CH_3O[/tex], obtained by finding the simplest whole-number ratio of the elements after dividing the number of atoms in the molecular formula by the greatest common divisor.
To determine the empirical formula of [tex]C_2H_6O_2[/tex], we need to find the simplest whole-number ratio of the atoms of each element in the compound. We can start by noting that the number of carbon (C), hydrogen (H), and oxygen (O) atoms as represented in the molecular formula [tex]C_2H_6O_2[/tex].
Looking at the subscripts of the elements, we can divide each by the greatest common divisor of the subscripts, which is 2. Doing this, we get:
Carbon: 2 / 2 = 1
Hydrogen: 6 / 2 = 3
Oxygen: 2 / 2 = 1
Therefore, the empirical formula is [tex]CH_3O[/tex]. To clarify, this is the simplest form of a formula representing the chemical composition.
As a check, if we consider the molecular formula of a very different compound, glucose, which is [tex]C_6H_{12}O_6[/tex], the empirical formula is [tex]CH_2O[/tex], because it is the simplest whole-number ratio, which in glucose's case is 1:2:1 for carbon, hydrogen, and oxygen atoms, respectively.
KAYLA BURNED A CANDLE IN A CLOSED SYSTEM WHERE MATTER CANNOT ENTER OR ESCAPE. GIVEN THIS SITUATION, WHAT SHOULD EQUAL THE MASS OF THE ORIGINAL CANDLE
Answer:same
Explanation:
Jason has five equally massed lead fishing weights, each weighing 250 grams. What is the total mass of the five leaf fishing weights
Answer:
Total mass of 5 lead fishing weights= 1250 grams
Explanation:
Given:
Mass of a lead fishing weight =250 grams
To find total mass of 5 such fishing weights
We can apply unitary method to find total mass of 5 fishing weights.
If mass of 1 lead fishing weight = 250 grams
∴ Total mass of 5 lead fishing weights = [tex]250\times 5 =1250\ grams[/tex]
1. Ir a mixture contains 0.639 g NaCl, 0.350 g NHACI, & 0.851 g of SiO2, with due regar
Signiricant figures, what is the mass percent of NaCl in the mixture?
34.73%
Explanation:We are given;
A mixture that contains;
NaCl = 0.639 g NH₄Cl = 0.350 g SiO₂ = 0.851 gWe are required to determine mass percent of NaCl in the mixture;
Step 1: Determine the total mass of the mixture;Mass of the mixture = Mass of NaCl + Mass of NH₄Cl + Mass of SiO₂
= 0.639 g + 0.350 g + 0.851 g
= 1.84 g
Step 2: Calculate the mass percent of NaCl in the mixtureThe mass % of NaCl = (Mass of NaCl ÷ Mass of the mixture) 100
= (0.639 g ÷ 1.84 g) × 100
= 34.73%
Therefore, the mass % of NaCl in the mixture is 34.73%
WILL MARK BRAINLIEST In the diagram, the black line represents the concentration of a reactant and the green line represents the concentration of a product.
Which statement best describes the reaction rate?
The reactants maintain an constant concentration in the first half of the reaction.
At the end of the reaction, both product and reactants are of an equal concentration.
The product maintains an constant concentration in the first half of the reaction.
At the end of the reaction, both product and reactants are of a constant concentration.
Answer:
slope of a line tangent to the product curve at any point in time
Explanation:
I believe the answers to the test are:
C
B
B
A
C
A
C
A
B
A
D
A
D
B
A
C
B
A
D
D
Answer:
At the end of the reaction, both product and reactants are of a constant concentration.
Explanation:
took the test
Condensation will most likely occur in a given volume
of air when the air is
A) saturated and contains no condensation nuclei
B) saturated and contains condensation nuclei
C) unsaturated and contains no condensation nuclei
D) unsaturated and contains condensation nuclei
Condensation is most likely to occur when air is saturated and contains condensation nuclei, as the saturation represents the maximum capacity for water vapor and the nuclei act as a base for the water vapor to condense into droplets.
Explanation:Condensation can most likely occur in a given volume of air when the air is saturated and contains condensation nuclei. The saturation of air refers to its maximum capacity to hold water vapor. After reaching its saturation point, air cannot dissolve more water vapor, and any further addition of water vapor will lead to condensation. On the other hand, condensation nuclei are tiny particles that water vapor latches onto to form water droplets, facilitating the process of condensation. Therefore, air that is not only saturated with water vapor but also contains condensation nuclei has the strongest propensity for condensation to occur.
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what happens when a certain number of hydrogen ions are mixed with the same number of hydroxide ions
Answer:
water is formed
Explanation:
acid and base maxing form salt and water by combining proton and hydroxyl ions
Which methods could be used to dilute a solution of sodium chloride (NaCl)?
Answer: Add more water to the solution and remove a small amount of the solution and mix it with water.
8. If I have 34 grams of FeO, how many grams of Oxygen were synthesized?
Fe + O2 → Feo
Answer:
34 gram of FeO produced 8 gram of oxygen.
Explanation:
Given data:
Mass of FeO = 34 g
Mass of oxygen = ?
Solution;
Chemical equation:
2FeO → 2Fe + O₂
Number of moles of FeO:
Number of moles = mass/ molar mass
Number of moles = 34 g /71.8 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of FeO with oxygen:
FeO : O₂
2 : 1
0.5 : 1/2 × 0.5 = 0.25
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 0.25 mol × 32 g/mol
Mass = 8 g
So 34 gram of FeO produced 8 gram of oxygen.
Vhat is the molarity of an Nal solution that contains 7.0 g of Nal in 23.0 mL of solution?
Answer:
Molarity = 5.22 M
Explanation:
Given data:
Mass of sodium chloride = 7.0 g
Volume of solution = 23.0 mL ( 23.0/1000 = 0.023 L)
Molarity = ?
Solution;
Number of moles of NaCl = 7.0 g/ 58.4 g/mol
Number of moles of NaCl = 0.12 mol
Molarity = moles of solute / volume in litter
Molarity = 0.12 mol / 0.023 L
Molarity = 5.22 M
For each of the following balanced chemical equations, calculate how many moles of product(s) would be produced if 0.500 mole of the first reactant were to react completely.
A) CO29(g) + 4h2(g) = CH4 (g) + 2H2O(l)
B) BaCl2 (aq) + 2AgANO3 (aq) = 2AgCl (s) + Ba(NO3)2 (aq)
c) C3H8 (g) + 5O2 (g) = 4H2O (l) + 3CO2(g)
d) 3H2SO4 (aq) + 2Fe (s) = Fe2(SO4)3 (aq) + 3H2 (g)
Answer:
A) 1.5 moles B) 1.5 moles C) 3.5 moles D) 2/3 moles
Explanation:
A) 1 mole of CO₂ gives 1 mole of CH₄ and 2 moles of H₂O, i.e, 3 moles of products. So 0.5 moles of CO₂ gives 3 x 0.5 = 1.5 moles of products.
B) 1 mole of BaCl₂ gives 2 moles of AgCl and 1 mole of Ba(NO₃)₂, i.e, 3 moles of products. So 0.5 moles of BaCl₂ gives 3 x 0.5 = 1.5 moles of products.
C) 1 mole of C₃H₈ gives 4 moles of H₂O and 3 moles of CO₂, i.e, 7 moles of products. So 0.5 moles of C₃H₈ gives 7 x 0.5 = 3.5 moles of products.
D) 3 moles of H₂SO₄ gives 1 mole of Fe₂(SO₄)₃ and 3 moles of H₂, i.e, 4 moles of products. So 0.5 moles of H₂SO₄ gives 4/3 x 0.5 = 2/3 moles of products.
To calculate the number of moles of product produced when a given amount of reactant is completely reacted, use the stoichiometric coefficients of the balanced chemical equation.
Explanation:To calculate the number of moles of product produced when a given amount of reactant is completely reacted, we need to use the stoichiometric coefficients of the balanced chemical equation. Each coefficient represents the ratio of moles between reactants and products.
For example, in the first equation, CO2 (g) + 4H2 (g) → CH4 (g) + 2H2O (l), the coefficient of CO2 is 1 and the coefficient of CH4 is also 1. This means that for every 1 mole of CO2 that reacts, 1 mole of CH4 is produced.
Therefore, if 0.500 mole of CO2 were to react completely, 0.500 mole of CH4 would be produced.
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Draw Lewis structures for each of the following.
1. nitrogen trifluoride, NF3
2. hydrogen sulfide, H2S
3. fluorine, F2
4. carbon monoxide, CO
5. sulfur dioxide, SO2
6. oxygen, O2
7. sulfur difluoride, SF2
8. boron trihydride, BHz
9. chloroform, CHCl3
10. carbon disulfide, CS2
11. beryllium chloride, BeCl2
12. hydrogen cyanide, HCN
13. acetylene, C2H2
14. silicon dioxide, SiO2
15. hydrogen peroxide, H2O2
16. sulfate, SO2-
17. methanol, CH3OH
18. nitrate, NO3
19. chlorite, CIO,
20. formic acid, CH2O
Answer:
Structures Attached
Explanation:
1. Total Number of e⁻ in NF₃
Number of e in Nitrogen = 5
Number of e in Flourine = 7
Total Number of e⁻ in NF₃ = 5 + 7(3)
Total Number of e⁻ in NF₃ = 5 + 21
Total Number of e⁻ in NF₃ = 26
2. Total Number of e⁻ in H₂S
Number of e in Hydrogen = 1
Number of e in sulphur = 6
Total Number of e⁻ in H₂S = 1(2) + 6
Total Number of e⁻ in H₂S = 2 + 6
Total Number of e⁻ in H₂S = 8
3. Total Number of e⁻ in F₂
Number of e in Flourine = 7
Total Number of e⁻ in F₂ = 7(2)
Total Number of e⁻ in F₂ = 14
4. Total Number of e⁻ in CO
Number of e in carbon = 4
Number of e in oxygen = 6
Total Number of e⁻ in CO = 4 + 6
Total Number of e⁻ in CO = 10
5. Total Number of e⁻ in SO₂
Number of e in Sulphur = 6
Number of e in oxygen = 6
Total Number of e⁻ in SO₂ = 6 + 6(2)
Total Number of e⁻ in SO₂ = 6 + 12
Total Number of e⁻ in SO₂ = 18
6. Total Number of e⁻ in O₂
Number of e in Oxygen = 6
Total Number of e⁻ in O₂ = 6(2)
Total Number of e⁻ in O₂ = 12
7. Total Number of e⁻ in SF₂
Number of e in Sulphur = 6
Number of e in Flourine = 7
Total Number of e⁻ in SF₂ = 6 + 7(2)
Total Number of e⁻ in SF₂ = 6 + 14
Total Number of e⁻ in SF₂ = 20
8. Total Number of e⁻ in BH₃
Number of e in Boron = 3
Number of e in Hydrogen = 1
Total Number of e⁻ in BH₃ = 3 + 1(3)
Total Number of e⁻ in BH₃ = 3 + 3
Total Number of e⁻ in BH₃ = 6
9. Total Number of e⁻ in CHCl₃
Number of e in Carbon = 4
Number of e in Hydrogen = 1
Number of e in chlorine = 7
Total Number of e⁻ in CHCl₃ = 4 + 1+ 7(3)
Total Number of e⁻ in CHCl₃ = 4 + 1 + 21
Total Number of e⁻ in CHCl₃ = 26
10. Total Number of e⁻ in CS₂
Number of e in Carbon = 4
Number of e in Sulphur = 6
Total Number of e⁻ in CS₂ = 4 + 6(2)
Total Number of e⁻ in CS₂ = 4 + 12
Total Number of e⁻ in CS₂ = 16
11. Total Number of e⁻ in BeCl₂
Number of e in Beryllium = 2
Number of e in Chlorine = 7
Total Number of e⁻ in BeCl₂ = 2 + 7(2)
Total Number of e⁻ in BeCl₂ = 2 + 14
Total Number of e⁻ in BeCl₂ = 16
12. Total Number of e⁻ in HCN
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Carbon = 4
Number of e⁻ in Nitrogen = 5
Total Number of e⁻ in HCN = 1 + 4 + 5
Total Number of e⁻ in HCN = 10
13. Total Number of e⁻ in C₂H₂
Number of e⁻ in Carbon = 4
Number of e⁻ in Hydrogen = 1
Total Number of e⁻ in C₂H₂ = 4(2) + 1(2)
Total Number of e⁻ in C₂H₂ = 8 + 2
Total Number of e⁻ in C₂H₂ = 10
14. Total Number of e⁻ in SiO₂
Number of e⁻ in Silicon = 4
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in SiO₂ = 4 + 6(2)
Total Number of e⁻ in SiO₂ = 4 + 12
Total Number of e⁻ in SiO₂ = 16
15. Total Number of e⁻ in H₂O₂
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in H₂O₂ = 1(2) + 6(2)
Total Number of e⁻ in H₂O₂ = 2 + 12
Total Number of e⁻ in H₂O₂ = 14
16. Total Number of e⁻ in SO₂⁻
Number of e in Sulphur = 6
Number of e in Oxygen = 6
Total Number of e⁻ in SO₂⁻ = 6 + 6(2)
Total Number of e⁻ in SO₂⁻ = 6 + 12
Total Number of e⁻ in SO₂⁻ = 18
17. Total Number of e⁻ in CH₃OH
Number of e⁻ in Carbon = 4
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in CH₃OH = 4 + 1(3) + 6 + 1
Total Number of e⁻ in CH₃OH = 4 + 3 + 6 + 1
Total Number of e⁻ in CH₃OH = 14
18. Total Number of e⁻ in NO₃
Number of e⁻ in Nitrogen = 5
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in NO₃ = 5 + 6(3)
Total Number of e⁻ in NO₃ = 5 + 18
Total Number of e⁻ in NO₃ = 23
19. Total Number of e⁻ in ClO
Number of e⁻ in Chlorine = 7
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in ClO = 7 + 6
Total Number of e⁻ in ClO = 7 + 6
Total Number of e⁻ in ClO = 13
20. Total Number of e⁻ in CH₂O
Number of e⁻ in Carbon = 4
Number of e⁻ in Hydrogen = 1
Number of e⁻ in Oxygen = 6
Total Number of e⁻ in CH₂O = 4 + 1(2) + 6
Total Number of e⁻ in CH₂O = 4 + 2 + 6
Total Number of e⁻ in CH₂O = 12
Lewis structures for different molecules are provided, including nitrogen trifluoride, hydrogen sulfide, fluorine, carbon monoxide, sulfur dioxide, oxygen, sulfur difluoride, boron trihydride, chloroform, carbon disulfide, beryllium chloride, hydrogen cyanide, acetylene, silicon dioxide, hydrogen peroxide, sulfate, methanol, nitrate, chlorite, and formic acid.
Explanation:Nitrogen trifluoride, NF3: Nitrogen (N) is the central atom and it forms single bonds with three fluorine (F) atoms. Each fluorine atom has a lone pair of electrons.Hydrogen sulfide, H2S: Hydrogen (H) is the central atom and it forms single bonds with two sulfur (S) atoms. Each sulfur atom has a lone pair of electrons.Fluorine, F2: Fluorine (F) forms a single bond with another fluorine atom.Carbon monoxide, CO: Carbon (C) is the central atom and it forms a triple bond with oxygen (O).Sulfur dioxide, SO2: Sulfur (S) is the central atom and it forms a double bond with one oxygen (O) atom and a single bond with another oxygen atom. The oxygen atom with the double bond has two lone pairs of electrons.Oxygen, O2: Oxygen (O) forms a double bond with another oxygen atom.Sulfur difluoride, SF2: Sulfur (S) is the central atom and it forms a single bond with two fluorine (F) atoms. The sulfur atom has two lone pairs of electrons.Boron trihydride, BH3: Boron (B) is the central atom and it forms three single bonds with hydrogen (H) atoms.Chloroform, CHCl3: Carbon (C) is the central atom and it forms a single bond with three hydrogen (H) atoms and a single bond with chlorine (Cl) atom. The chlorine atom has three lone pairs of electrons.Carbon disulfide, CS2: Carbon (C) is the central atom and it forms double bonds with two sulfur (S) atoms.Beryllium chloride, BeCl2: Beryllium (Be) is the central atom and it forms two single bonds with chlorine (Cl) atoms.Hydrogen cyanide, HCN: Carbon (C) is the central atom and it forms a triple bond with nitrogen (N). The nitrogen atom has a lone pair of electrons.Acetylene, C2H2: Carbon (C) is the central atom and it forms triple bonds with another carbon atom. Each carbon atom forms single bonds with two hydrogen (H) atoms.Silicon dioxide, SiO2: Silicon (Si) is the central atom and it forms a double bond with one oxygen (O) atom and a single bond with another oxygen atom. The oxygen atom with the double bond has two lone pairs of electrons.Hydrogen peroxide, H2O2: Oxygen (O) is the central atom and it forms single bonds with two hydrogen (H) atoms. Each oxygen atom has two lone pairs of electrons.Sulfate, SO2-: Sulfur (S) is the central atom and it forms a double bond with one oxygen (O) atom and two single bonds with other oxygen atoms. The oxygen atoms with single bonds have a lone pair of electrons.Methanol, CH3OH: Carbon (C) is the central atom and it forms single bonds with three hydrogen (H) atoms, a single bond with oxygen (O) atom, and also has a lone pair of electrons. The oxygen atom has two lone pairs of electrons.Nitrate, NO3-: Nitrogen (N) is the central atom and it forms a double bond with one oxygen (O) atom and two single bonds with other oxygen atoms. The oxygen atoms with single bonds have a lone pair of electrons.Chlorite, CIO-: Chlorine (Cl) is the central atom and it forms a single bond with oxygen (O) atom and has two lone pairs of electrons. The oxygen atom has two lone pairs of electrons.Formic acid, CH2O: Carbon (C) is the central atom and it forms single bonds with two hydrogen (H) atoms, a single bond with oxygen (O) atom, and also has a lone pair of electrons. The oxygen atom has two lone pairs of electrons.Learn more about Lewis structures here:https://brainly.com/question/34631490
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How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?
Answer:
[tex]\large \boxed{\text{77.4 mL}}[/tex]
Explanation:
Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O
V/mL: 249
c/mol·L⁻¹: 0.0443 0.285
1. Calculate the moles of Ba(OH)₂
[tex]\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}[/tex]
2. Calculate the moles of HCl
The molar ratio is 2 mol HCl:1 mol Ba(OH)₂
[tex]\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}[/tex]
3. Calculate the volume of HCl
[tex]V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}[/tex]
You burn a piece of paper. After burning the paper, you find the mass of the remaining ash. The mass of the ash does NOT equal the mass of the paper you started with. How can you account for the different in mass? Where could it have gone?
Which process takes place when large rocks crumble due to oxidation?
A) Erosion
B) Deposition
C) physical weathering
D) chemical weathering
Answer:
chemical weathering I think it is a right answers
4. 450g of chromium (III) sulfate reacts with excess potassium phosphate. How many grams of
potassium sulfate will be produced? (ANS: 6.0x10ʻg K2SO4)
Answer:
[tex]1.6\times 103[/tex]g of K₂SO₄ will be produced
Explanation:
Given [tex]K_3PO_4[/tex] is available in excess
Reaction:
Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄
From the above reaction, it is clear that the moles of (Cr₂(SO₄)₃), [tex]K_3PO_4[/tex] and K₂SO₄ are 1, 2 and 3 respectively
We can say that 1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄
molar mass of Cr₂(SO₄)₃ = 147 g/mol
molar mass of K₃PO₄ = 212 g/mol
molar mass of K₂SO₄ = 174 g/mol
We can write as;
Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄
1 mol (147 g/mol) 2 mol (212 g/mol) 3 mol (174g/mol)
Therefore, we have
Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄
147 g 424 g 522 g
So, we can see that 147 g of Cr₂(SO₄)₃ reacts with 424 g of 2K₃PO₄ to produce 522 g of K₂SO₄
147 g of Cr₂(SO₄)₃ = 522 g of K₂SO₄
So, 450 g of Cr₂(SO₄)₃ = [tex]\frac{(522\times 450)}{147}[/tex]g of K₂SO₄ = 1597.959 g = [tex]1.59\times 103[/tex] g = [tex]1.6\times 103[/tex] g
So,[tex]1.6\times 103[/tex]g of K₂SO₄ will be produced
Two types of heterogeneous mixtures are suspensions and colloids.
true or false
Explanation:
Heterogeneous mixture is a mixture with non-uniform composition.
The properties of the mixture like concentration may change for different parts of the mixture.
Colloids contain solute particles of size [tex]2nm-500nm[/tex].The presence of these particles makes the mixture heterogeneous.
Suspensions contain solute particles of size [tex]500nm-1000nm[/tex].These particles settle to the bottom of the mixture which makes the composition of the bottom different from the top.
So,colloids and suspensions are two types of heterogeneous mixtures.
Final answer:
Yes, the statement is true; suspensions and colloids are both types of heterogeneous mixtures, differentiated by the size of their particles and their tendency to settle.
Explanation:
The statement is true: suspensions and colloids are indeed two types of heterogeneous mixtures. A suspension is a heterogeneous mixture where large particles are dispersed throughout a liquid or gas and can settle over time, examples include paint and mud. On the other hand, a colloid is also a heterogeneous mixture, but it consists of much finer particles that don't settle upon standing, such as fog, milk, and butter.
Colloids can be found in various states like sols (solid particles in a liquid), gels (solids absorbing liquids), aerosols (solid or liquid particles in a gas), or emulsions (liquids dispersed in other liquids). They exhibit the Tyndall effect, where light is scattered by the particles, distinguishing them from solutions, which are homogenous and clear. Suspensions, with their larger particle size, do not maintain their dispersion upon standing and are often cloudy.
It's important to note that while suspensions and colloids are heterogeneous due to the presence of dispersed particles, these particles differ significantly in size between the two types, which influences the stability and appearance of the mixture.
if 115 G of a substance reacts with 84 grams of another substance what will be the mass of the products after the reaction
Answer:
199 g
Explanation:
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
For example:
In given reaction there are two reactant one with a mass of 115 g and other with the mass of 84 g thus the resultant product must have a mass of 199 g.
Chemical equation:
A + B → AB
115 g + 84 g = 199 g
119 g = 199 g
Final answer:
The mass of the products after the reaction between 115 grams and 84 grams of two substances, according to the Law of Conservation of Mass, will be 199 grams, assuming no material is lost during the reaction.
Explanation:
If 115 grams of a substance reacts with 84 grams of another substance, according to the Law of Conservation of Mass, the mass of the products after the reaction will be the sum of the masses of the reactants. This is because in a chemical reaction, matter is neither created nor destroyed, so the total mass of the reactants must equal the total mass of the products.
In this case, adding the masses of the reactants gives us:
115 g + 84 g = 199 g
Therefore, the mass of the products after the reaction will be 199 grams. However, this assumes a perfectly efficient reaction with no losses to factors such as gas escaping the reaction vessel or side reactions. In a practical setting, a slight discrepancy might be observed due to such factors.
The procedure for an experiment is _____.
A. the variable that you will change
B. a set of safety rules that you need to follow in the lab
C. what you're going to do and how you're going to do it
D. research you've done about the topic you're investigating
Answer:
the correct answer would be c.
Answer: The correct answer is: C. what you're going to do and how you're going to do it
Explanation: A good experiment should follow a design with a few logical steps. An experiment must be carried out methodically in order to perform valid and reliable measurements. Experimenting is a procedure to check one or several hypotheses about a specific topic.
Write the equations for the esterification process between ethanoic acid and 2-
propanol with sulphuric acid as catalyst.
Answer: CH3COOH + CH3CHOHCH3 -> CH3COOCH(CH3)CH3 + H2O
Explanation: Estherification is the reaction between alkanoic acids and alkanols to porduce esters and what. ethanoic acid will react with 2-propanol to produce an ester and water.
Ethanoic acid reacts with 2-propanol to form an ester and water in the presence of sulfuric acid.
[tex]\bold {CH_3COOH + CH_3CHOHCH_3 \rightarrow CH_3COOCH(CH_3)CH_3 + H_2O}[/tex]
What is Esterification?It is a type of reaction in which typically acid and alcohol react to form an ester product.The ester is derived from acid when the hydroxyl group is replaced by an alkyl group.Ethanoic acid reacts with 2-propanol to form an ester and water in the presence of sulfuric acid.[tex]\bold {CH_3COOH + CH_3CHOHCH_3 \rightarrow CH_3COOCH(CH_3)CH_3 + H_2O}[/tex]
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Which substance is a mixture?
table salt
gasoline
aluminum
carbon dioxide
Answer:
Gasoline
Explanation:
Pure Substance: are those having same type of atoms or molecules.
Compound: one or more atoms of the element combine to form a compound.
Mixture: is combination of one or more substances mix together.
Homogeneous Mixture: are those which are not distinguish by naked eye but can be separated into its components by physical means
Heterogeneous mixture: are those which can be distinguishing by naked eye and can be separated into its components by physical means.
So keeping in mind the above definitions
Table salt:
The chemical formula of table salt is NaCl and it a pure compound not mixture.
Gasoline:
Gasoline is mixture of hydrocarbon, It contains small hydrocarbons ranging from 4 Carbon to 12 Carbon per molecule. it is a homogeneous mixture.
Aluminum:
Aluminum is a pure substance that is made up of same kind of atoms. so it is an element and not mixture. Its symbol is Al
Carbon dioxide:
Carbon dioxide is a pure compound and its chemical formula is CO₂ and not a mixture.
So, the write option is Gasoline.
Answer:
gasoline
Explanation:
Please help! Thank you ❤️
Answer:
All of these
option a to f are molecule
Explanation:
Molecule:
Molecule is formed when two or more than two atoms of an element combine chemically.
N₂O₄ is a molecule where Nitrogen combine with oxygen through covalent bond.
The name of this compound is Dinitrogen tetraoxide.
Na₂CO₃ is a Compound where sodium carbon and oxygen combined to form a Compound.
Its name is sodium carbonate.
CClF3 is a Compound: It is a polar Compound
It named as Chlorofloromethan.
OBr₂ is polar Molecule. This molecule consist of oxygen and Bromine atoms and a covalent compound.
It named as oxygen dibromide.
FeSO₄ is a compound . it consists of iron Oxygen and sulfate atoms.
It named as Iron Sulfate.
LiBr is molecule. it is ionic in nature and Known as molecular compound
It named as Lithium bromide.
All molecule can not be compound as some molecule have same type of atom such as ozone or hydrogen molecule., but all compounds are molecule, so with this concept all of these are molecule.
A beam of light has a wavelength of 24.0 μm ( μm = 10 -6m). What is its frequency?
Choices:
a. 1.25 x 1013 Hz
b. 8.00 x 10-8 Hz
c. 1.25 x 107 Hz
d. 8.00 x 10-11 Hz
Answer:
Frequency = 1.25 ×10¹³ Hz
Explanation:
Given data:
Wavelength of light = 24.0 μm (2.4 ×10⁻⁵ m)
Frequency = ?
Solution:
Formula:
Speed of light = wavelength × frequency
Speed of light /wavelength = frequency
Frequency = 3×10⁸ m /s /2.4 ×10⁻⁵m
Frequency = 1.25 ×10¹³ s⁻¹
s⁻¹ = Hz
Frequency = 1.25 ×10¹³ Hz
Which of the following is true of gas?
A gas is not compressed because of the low intermolecular forces.
a gas does not easily mix with other gases becasye of the high density of substances as gases.
gas particles have high intermolecular forces, leading to the ability to flow easily. gas particles spread out to fill a container, leading to a low density of gas
Answer:
Gases mix easily because of their high kinetic energy and low inter-molecular forces.
Explanation:
Answer:
Gases mix easily because of their high kinetic energy and low inter-molecular forces.
Explanation:
Please help will give 40 points and brainiest
Examine the following reactions.
Reaction 1
PCl3(g) + Cl2(g) → PCl5(g)
Reaction 2
PCl5(g) → PCl3(g) + Cl2(g)
Which options correctly explain the relationship between the two reactions?
Select all that apply.
Both reactions contain entirely gaseous substances, making them both decomposition reactions.
Both reactions contain all the same substances, making them both synthesis reactions.
Reaction 1 is the opposite of Reaction 2.
The reactant in Reaction 2 breaks apart, making it a decomposition reaction.
The reactants in Reaction 1 combine instead of breaking apart, making it a synthesis reaction.
Reaction 1 is the same as Reaction 2.
The relationship between the two reactions is correctly explained as chemical equation 1 is the opposite of Reaction 2.The reactant in Reaction 2 breaks apart, making it a decomposition reaction.
What is chemical equation?
Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.
A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .
The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.
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Create a story and describe the motion (in meters) of an object that has traveled a distance but has no displacement. Need Help ASAP
Answer:
One complete revolution around a circular path.
Explanation:
Let us take the case of a car moving in a circular track of radius r metres.
In one revolution, the car covers the length(distance) equal to the perimeter of the circle.
In this case, distance traveled = 2[tex]\pi[/tex]r metres
But after one complete revolution, the car reaches the same position as it was at the beginning of the motion.
Hence, the initial and final points coincide or the car hasn't changed it's position w.r.t the initial point.
So in this case, the displacement is zero.
Hence, revolution of a car around a circular path is an example of an object traveling a distance but having no displacement.
(Temperature is negative) How many liters of NO2 (at STP) can be produced from 149 grams of Cu reacting with concentrated nitric acid, HNO3 (aq)?
(The equation is balanced)
Cu + 4 HNO3 ⇒ Cu(NO3) 2 + 2 H2O + 2 NO2
63.55, 63.02, 187.57, 18.02, 46.01
^ g/mol
Answer:
[tex]\large \boxed{\text{106 L}}[/tex]
Explanation:
We will need a balanced chemical equation with molar masses and volumes, so, let's gather all the information in one place.
MV/L: 22.71
M_r: 63.55
Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂
m/g: 149
(a) Moles of Cu
[tex]\text{Moles of Cu } =\text{149 g Cu } \times \dfrac{\text{1 mol Cu }}{\text{63.55 g Cu }} =\text{2.344 mol Cu}[/tex]
(b) Moles of NO₂
The molar ratio is 2 mol NO₂:1 mol Cu
[tex]\text{Moles of NO$_{2}$}= \text{2.344 mol Cu} \times \dfrac{\text{2 mol NO$_{2}$}}{ \text{1 mol Cu}} = \text{4.689 mol NO$_{2}$}[/tex]
(c) Volume of NO₂
The volume of 1 mol of an ideal gas at STP (0 °C and 1 bar) is 22.71 L.
[tex]\text{V} = \text{4.689 mol} \times \dfrac{\text{22.71 L}}{\text{1 mol}} = \textbf{106 L}\\\\\text{You can produce $\large \boxed{\textbf{106 L}} $ of NO$_{2}$.}[/tex]
If 518.0 g sample of a metal has a volume of 2.00mL what is the density of the metal
Answer:
Density of metal = 259 g/mL
Explanation:
Data Given:
Mass of metal = 518.0 g
Volume of Sugar = 2.00 mL
Density of = ? g/mL
The formula will be used is
d = m/v ........................................... (1)
where
d is density
m is the mass
v is the volume
So
put the given values in Equation (1)
d = 518.0 g / 2.00 mL
d = 259 g/mL
Density of metal = 259 g/mL