Answers
Answer:
KOH
Explanation:
Chemical reaction:
2K + 2H₂O → 2KOH + H₂
Element ratio of K.
K = 1
Because only potassium is present.
Element ratio of H₂O.
2 : 1
in water ratio of element is 2 : 1 because two hydrogen and one oxygen atom present.
Element ratio of KOH
1 : 1 : 1
in KOH elemental ratio is 1 : 1 : 1 because one potassium one hydrogen and one oxygen atom are present.
Element ratio of H₂.
2
Just two atoms of hydrogen are present.
Related Questions
How many decigrams are in 4.6 decagrams?
0.0046 decigrams
0.046 decigrams
460 decigrams
4,600 decigrams
Answers
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]
Please help would really appreciate it! Thank you so much! ❤️ Have a wonderful day!
Answers
Answer:
option b : O₉S₁₀
Explanation:
There are some points to name a covalent compound, according to that
number of atoms numbered as
**mono for one, di - for two tri for three and so on.
the name of the atom as
**carbon as carbide, sulfur as sulfide, oxygen as oxide, foulrine as flouro etc.
So,
Formula of Nonoxide decasulfide
Now Naming of the Chemical compound
Following are the explanations of terms
Prefix nono is used for number ''9'' denoting to the number of atomOxide used for oxygendeca is use for number ''10'' denoting to the number of atomSulfide is used for Sulfur in a compoundSo keeping the above points in mind
The right answer is
Option b : O₉S₁₀
How could soil and a mouse interact in an ecosystem?
Answers
Answer:
Explanation:
It could make burrows and homes in the dirt. It could hide food and itself in the soil. The Soil will grow food.
In an ecosystem , a mouse would make burrows in soil and soil would provide nutrients to the mouse.
What is an ecosystem?Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.
Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.
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Bob mixed 5 mg of sodium hydroxide pellets in 100 mL of water. Describe how the rate of the solution changes when he adds 1 mg more of sodium hydroxide pellets.
Answers
Answer:
As Bob add more solute to dissolves, the rate of solution will be decreases
Explanation:
At certain temperature the maximum amount of solute dissolve in the specific amount of solution or solvent.
There are many factors that affect solubility of the solute in solvent. Nature of the solute also affects the rate of solubility.
Temperature also affects the rate of solubility. Solubility increase when the temperature increases.
Pressure: Change in pressure have no effect on the rate of solubility
The rate of solution: is a measure that how fast a substance dissolves in solvent.
Amount of solute already dissolved affect the rate of solution. If a solvent have little amount of solute in the solution it dissolve quickly. But when you have more solute in the solution the process was slower.
1 g of lead dissolves in 100 g of water at room temperature. Similarly 5 mg of sodium hydroxide pellet dissolve in 100ml of water at room temperature.
So if Bob add 1 mg more of NaOH in the same 100ml of water the rate of the solubility decreases.
As there are some factors that decrease the solubility rate of NaOH pellet as follow:
• large size of the solute dissolve slowly as the solubility process take on the surface of the solute particles, powders dissolve fast.
• less amount of solvent with ratio to solute
• saturated solution
So due to all above reasons the rate of the solubility decreases upon adding a little more amount of NaOH pellet.
Answer:
a
Explanation:
:)
u are given a 30.0g sample of radioactive uranium that has a half-life of 12 hours. How much uranium will be left after 2 days? 1Choice 1 3.75g 2Choice 2 15.0g 3Choice 3 60.0g 4Choice 4 1.88g
Answers
1.88 g
Explanation:We are given;
Mass of a radioactive sample of Uranium as 30.0 g Half life of Uranium sample is 12 hoursTime is 2 daysWe are required to determine the amount of uranium left after 2 days
We are going to use the formula;N = N₀ × 0.5^n
Where, N is the remaining mass, N₀ is the original mass and n is the number of half lives.
n = (2 × 24) ÷ 12 hours
= 4
Therefore;
Remaining mass, N = 30.0 g × 0.5^4
= 1.875 g
= 1.88 g
Therefore, the mass of the remaining sample after decay is 1.88 g
How can general properties influence element location on a periodic table?
Answers
Answer:
The elements in the periodic table are arranged in order of increasing atomic number.
Explanation:
The elements are arranged based on their atomic numbers in a periodic table, not based on their atomic masses.
• The chemical elements in the periodic table are demonstrated in order of their atomic number.
• In the modern periodic table, the arrangement of elements is done based on their atomic number, not their relative atomic masses.
• In the periodic table, the horizontal rows, known as periods, show the arrangement of elements in order of increasing atomic number.
Thus, based on atomic numbers, elements are arranged in a periodic table.
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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)
Answers
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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Write the equations for the esterification process between ethanoic acid and 2-
propanol with sulphuric acid as catalyst.
Answers
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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Vhat is the molarity of an Nal solution that contains 7.0 g of Nal in 23.0 mL of solution?
Answers
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
4. 450g of chromium (III) sulfate reacts with excess potassium phosphate. How many grams of
potassium sulfate will be produced? (ANS: 6.0x10ʻg K2SO4)
Answers
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
What's C2H6O2 in empirical formula?
Answers
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.
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
Answers
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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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.
Answers
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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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
Answers
Answer:same
Explanation:
if 115 G of a substance reacts with 84 grams of another substance what will be the mass of the products after the reaction
Answers
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
Answers
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.
How many moles are there in 8.50 X 10^24 molecules of sodium sulfate, Na2SO3?
(Molecules to Moles)
A.) 51.2 moles
B.) 5.12x10^48 moles
C.)14.1 moles
D.)1.40 Moles
Answers
Answer:
C) 14.1 moles
Explanation:
1 mole = 6.02 *10^ 23
8.50 * 10^24 molecules *(1 mole/6.02*10^23 molecules) ≈ 14.1 moles
The number of moles there are in 8.50 X [tex]10^{24}[/tex] molecules of sodium sulfate is 14.1 moles (option C).
How to calculate number of moles?The number of moles in a molecule of a substance can be calculated by dividing the number of molecules by Avogadro's number as follows:
no of moles = no of molecules ÷ 6.02 × [tex]10^{23}[/tex] molecules
According to this question, there are in 8.50 X [tex]10^{24}[/tex] molecules of sodium sulfate in a substance. The number of moles in this substance can be calculated as follows:
no of moles = 8.50 X [tex]10^{24}[/tex] ÷ 6.02 × [tex]10^{23}[/tex]
no of moles = 14.1 moles
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Find both the volume percent of a solution that has 10.0 g of ethanol (D = 0.7893 g/mL) and 90.0 g of water (D = 0.9987 g/mL). Assume volumes are additive.
Answers
Answer:
Volume percent of a solution that has 10.0 g of ethanol= 12.37%
Volume percent of a solution that has 90.0 g of water = 87.69%
Explanation:
Volume Percent:
The volume percent helps us to indicate the concentration of the solution when the volume of the solute and volume of the solution is given by
[tex]Volume Percent=\frac{\text{Volume of the Solute}}{\text{Volume of Solution}}\times100%[/tex]
With respect to density 1mL of ethanol
=> 0.7893g or 10g of methanol contains
=> [tex]\frac{1}{0.7893}\times10[/tex]
=> 12.67mL
And for water = [tex]\frac{1}{0.9987} \times 90.0[/tex]
=> 90.12mL
Now total volume = 90.12+ 12.67 = 102.79mL
%(v/v) for water = [tex]\frac{90.12}{102.79}\times 100[/tex]
= 87.69%
%(v/v) for ethanol = [tex]\frac{12.67}{102.79}\times 100[/tex]
= 12.37%
The volume percent of ethanol in the solution is 12.3%, and the volume percent of water is 87.7%. This is calculated by determining the volume of each substance using their respective densities and the mass provided and then finding the percentage of each substance's volume in the total solution volume.
Explanation:To find the volume percent of a solution that has 10.0 g of ethanol with a density (D) of 0.7893 g/mL, and 90.0 g of water with a density (D) of 0.9987 g/mL, we first calculate the volumes of each component. For ethanol, divide 10.0 g by 0.7893 g/mL to get the volume in milliliters. For water, divide 90.0 g by 0.9987 g/mL to get the volume in milliliters.
To calculate the volume of ethanol: 10.0 g / 0.7893 g/mL = 12.7 mL
To calculate the volume of water: 90.0 g / 0.9987 g/mL = 90.2 mL
Assuming volumes are additive: Total volume of the solution = 12.7 mL (ethanol) + 90.2 mL (water) = 102.9 mL
Now, we can find the volume percent of ethanol: (volume of ethanol / total volume of solution) × 100% = (12.7 mL / 102.9 mL) × 100% = 12.3%
The volume percent of water is similarly calculated: (volume of water / total volume of solution) × 100% = (90.2 mL / 102.9 mL) × 100% = 87.7%
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determine the number of cobalt atoms in a 117.86 g sample of cobalt
Answers
Answer:
12.04 × 10²³ atoms
Explanation:
Given data:
Mass of cobalt = 117.86 g
Number of atoms = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
Number of moles of cobalt = 117.86 g/ 58.9 g/mol
Number of moles of cobalt = 2 mol
one moles of cobalt = 6.022 × 10²³ atoms of cobalt
2 moles × 6.022 × 10²³ atoms / 1 mol
12.04 × 10²³ atoms
Calculate the amount of CuSO4•5H2O required to make 25 mL of a 0.45 M solution of CuSO4.
Answers
Answer:
Mass = 1.76 g
Explanation:
Given data:
Volume of solution = 25 mL
Molarity = 0.45 M
Amount of CuSO₄.5H₂O = ?
Solution:
Molarity = number of moles / volume in litter
Number of moles = Molarity × Volume in L
Number of moles = 0.45 M × 0.025 L
Number of moles = 0.011 mol
Mass of f CuSO₄.5H₂O:
Mass = Number of moles × molar mass
Mass = 0.011 mol × 159.6 g/mol
Mass = 1.76 g
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?
Answers
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]
Oxalic acid reacts with sodium hydroxide according to the following equation:
H2C2O4 + 2 NaOH → Na2C20. + 2 H20
A 25.00 mL sample of the H2C2O solution required 19.62 mL of 0.341 M NaOH for
neutralization. Calculate the molarity of the acid.
Answers
Answer:
Concentration = 0.14 M
Explanation:
Given data:
Volume of base = 19.62 mL
Molarity of base = 0.341 M
Volume of acid = 25.00 mL
Molarity of acids = ?
Solution:
Number of moles of base = Molarity × volume in litter
Number of moles = 0.341 mol/L × 0.02 L
Number of moles = 0.007 mol
Concentration of acid:
Concentration = 1/2 (0.007 mol) / 25 mL × 10⁻³ L/mL
Concentration = 0.0035 mol / 25 × 10⁻³ L
Concentration = 1.4 × 10⁻¹ M
Concentration = 0.14 M
Cathode rays are composed of_____.
a. protons
b. electrons
c. neutrons
Answers
Answer:
negatively-charged particles.(ELECTRON) B
Explanation:
Electrons are the negatively charged particles of atom. Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus
Jason has five equally massed lead fishing weights, each weighing 250 grams. What is the total mass of the five leaf fishing weights
Answers
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]
can someone please please explain how to write the formula for a compound containing a polyatomic ion. If you could include examples that would be great too!
Answers
Answer:
Explanation:
Polyatomic ions are ions which consist of more than one atom. For example, nitrate ion, NO3-, contains one nitrogen atom and three oxygen atoms. The atoms in a polyatomic ion are usually covalently bonded to one another, and therefore stay together as a single, charged unit.
To write the formula for a compound with a polyatomic ion, consider that the compound should be electrically neutral. Balance the charges from different ions, treating polyatomic ions as discrete units. For instance, calcium phosphate (Ca3(PO4)2) illustrates this with three calcium ions and two phosphate ions.
Explanation:When writing the formula for a compound containing a polyatomic ion, you must consider that the compound has to be electrically neutral overall. This often involves balancing the number of positive and negative charges coming from different ions. The polyatomic ions are usually treated as discrete units in the formula.
For example, consider calcium phosphate, which has the formula Ca3(PO4)2. Here, there are three calcium ions (Ca²+), and two phosphate groups (PO4³-). Each PO4³- group comprises one phosphorus atom and four oxygen atoms. The overall compound is electrically neutral because the positive charges from the three calcium ions balance out the negative charges from the two phosphate ions.
Another example can be seen in the ionic compound sodium oxalate, which has the formula Na₂C₂O₄. Here, the polyatomic ion is C₂O₄²-, and the compound has two sodium ions (Na⁺) for each oxalate ion. The formula of this compound is not empirical because it represents the number of atoms in the discrete unit of the oxalate ion.
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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
Answers
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:
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
Answers
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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Which chemical equation represents a precipitation reaction?
Answers
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 moles are in 75.02 grams of strontium
Answers
Answer:
1 grams Strontium Fluoride to mol = 0.00796 mol
10 grams Strontium Fluoride to mol = 0.07961 mol
50 grams Strontium Fluoride to mol = 0.39804 mol
100 grams Strontium Fluoride to mol = 0.79607 mol
200 grams Strontium Fluoride to mol = 1.59214 mol
500 grams Strontium Fluoride to mol = 3.98036 mol
1000 grams Strontium Fluoride to mol = 7.96072 mol
- Hopefully this helps a bit :)
Which are examples of physical weathering? Check all that apply.
grinding away of rock surface
top layers flaking off due to release of pressure
water dissolving soft rock
rust forming due to oxidation
temperature changes from hot to cold
animals digging burrows to make homes
water freezing, expanding and thawing repeatedly
acid rain weathering a statue
tree roots growing in the cracks of rock
Answers
Final answer:
Physical weathering refers to the process of breaking down rocks without changing their chemical composition. Examples include grinding away of rock, flaking due to pressure release, and water freezing and thawing.
Explanation:
Physical weathering refers to the process of breaking down rocks into smaller pieces without changing their chemical composition. Examples of physical weathering include the grinding away of a rock surface, the top layers of rock flaking off due to the release of pressure, and the freezing and thawing of water, which causes it to expand and contract and break the rock apart.
Other examples include tree roots growing in the cracks of rocks and animals digging burrows. On the other hand, rust forming due to oxidation and acid rain weathering a statue are examples of chemical weathering.
Final answer:
Physical weathering includes processes like abrasion, exfoliation, temperature fluctuations, freeze-thaw cycles, animal activities, and root wedging, all of which disintegrate rock mechanically.
Explanation:
The examples of physical weathering are those processes that result in the breakdown of rock without altering its chemical composition. Here are the examples that correctly apply to the concepts of physical weathering:
Grinding away of rock surface due to abrasionTop layers flaking off due to release of pressure (exfoliation)Temperature changes from hot to cold causing expansion and contractionWater freezing, expanding, and thawing repeatedly in cracks (freeze-thaw weathering)Animals digging burrows to make homes (biomechanical weathering)Tree roots growing in the cracks of rock (root wedging)On the other hand, the following are not examples of physical weathering because they involve chemical changes to the rock: water dissolving soft rock, rust forming due to oxidation, and acid rain weathering a statue.