Predict the products of the reaction below. That is, complete the right-hand side of the chemical equation. Be sure your equation is balanced and contains state symbols after every reactant and product.

HNO3 (aq)+ H2O â_________

The new volume : 21.85 ml

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

V2

Solution

Combined gas law :

[tex]\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}[/tex]

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

Answer:

The Aluminium sample

Explanation:

From the question,

ΔQ of the iron = Cm(t₂-t₁)........................ Equation 1

Given: C = specific heat capacity of iron, m = mass of iron, t₁ and t₂ = initial and final temperature respectively.

Given: m = 175 g = 0.175 kg, t₂ = 21.5°C, t₁ = 99.7°C

Constant: C = 444 J/kgK.

Substitute into equation 1

ΔQ = 0.175(444)(21.5-99.7)

ΔQ = -6076.14 J

Similarly, for aluminium,

ΔQ' = c'm'(t₂-t₁)...................... Equation 2

Given: m' = 175 g = 0.175 kg,

Constant: 900 J/kgK

ΔQ' = 0.175(900)(21.5-99.7)

ΔQ' = -12316.5 J

Hence the aluminium sample undergoes the greater heat change

Answer:

11.648 g

Explanation:

From the question,

Note: an object immersed in a fluid, displaced an amount of fluid, equal to its volume.

Density = mass/volume.

D = M/V................................................................ Equation 1

Where D = density of the copper block, M = mass of the copper block, V = volume of the copper block.

make M the subject of the equation

M = D×V......................................... Equation 2

Given: D = 8.96 g/mL, V = 1.3 mL

Substitute into equation 2

M = 8.96×1.3

M = 11.648 g

Hence the mass of the block is 11.648 g

Answer:

[tex]20[/tex] moles of CO2 can be produced from a reaction of 10.0 moles C2H6

Explanation:

In this reaction -

2 moles of C₂H6 produces four molecules of Carbon dioxide (CO2)

So 1 mole of C₂H6 will produce [tex]\frac{4}{2} = 2[/tex] moles of Carbon dioxide (CO2)

Thus, 10 moles of C₂H6 will produce [tex]2 * 10 = 20[/tex] moles of Carbon dioxide (CO2)

Answer:

The object is moving at a constant speed.

Mass of Na₂SO₄= 14.346 g

Given

17 g NaHCO₃

0.4 mol H₂SO₄

Required

mass of Na₂SO₄

Solution

Reaction

H₂SO₄ (aq) + 2 NaHCO₃ (aq) ⇒ Na₂SO₄(aq) + 2 CO₂(g) + 2 H₂O (l)

mol NaHCO₃ :

= mass : MW

= 17 g : 84 g/mol

= 0.202

Find limiting reactant = mol : coefficient

H₂SO₄ = 0.4 :  1 = 0.4

NaHCO₃ = 0.202 : 2 = 0.101

NaHCO₃ as a limiting reactant (smaller ratio)

Mol Na₂SO₄ based on mol NaHCO₃

From the equation, mol ratio NaHCO₃ : Na₂SO₄ = 2 : 1, so mol Na₂SO₄ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.202

= 0.101

Mass of Na₂SO₄ :

= mol x MW

= 0.101 x 142,04 g/mol

= 14.346 g

Answer:

[tex]r_{H_2} = \frac{-1}{2} r_{HI}[/tex]

Explanation:

Hello!

In this case, considering the given chemical reaction:

[tex]H_2(g) + I_2(g) \rightarrow 2HI(g)[/tex]

Thus, by applying the law of rate proportions, we can write:

[tex]\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}[/tex]

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

[tex]r_{H_2} = \frac{-1}{2} r_{HI}[/tex]

Best regards!

Answer:

All organisms begin life as a single cell

Explanation:

Single cell also did you know 15 minutes can save you 15% more on car insurance

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. [tex]Na_{2} CO_{3}[/tex]

Major Species at B:

1. [tex]Na_{2} CO_{3}[/tex]

2. [tex]NaHCO_{3}[/tex]

Major Species at C:

1. [tex]NaHCO_{3}[/tex]

Major Species at D:

1. [tex]NaHCO_{3}[/tex]

2. [tex]H_{2}CO_{3}[/tex]

Major Species at E:

1. [tex]H_{2}CO_{3}[/tex]

Major Species at F:

1. [tex]H_{2}CO_{3}[/tex]

b) pH calculation:

At Halfway point B:

pH = pK[tex]a_{1}[/tex] + log[[tex]CO_{3}[/tex][tex].^{-2}[/tex]]/[H[tex]CO_{3}[/tex][tex].^{-1}[/tex]]

pH = pK[tex]a_{1}[/tex] = 6.35

Similarly, at halfway point D.  

At point D,

pH = pK[tex]a_{2}[/tex] + log [H[tex]CO_{3}[/tex][tex].^{-1}[/tex]]/[H2[tex]CO_{3}[/tex]]

pH = pK[tex]a_{2}[/tex] = 10.33

Answer:

Answer:A

Answer:AExplanation:

Answer:AExplanation:Molar Mass of glucose = (6×12)+(1×12)+(16×6)= 180g/mol

= 180g/molNumber of moles of Glucose = Mass/Molar Mass

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556= 2555.55

Answer:

Answer:A

Answer:AExplanation:

Answer:AExplanation:Molar Mass of glucose = (6×12)+(1×12)+(16×6)= 180g/mol

= 180g/molNumber of moles of Glucose = Mass/Molar Mass

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556= 2555.55

Answer:

0.0802 M.

Explanation:

Hello!

In this case, since the barium acetate has the following formula:

[tex]Ba(CH_3COO)_2[/tex]

So its molar mass is 255.43 g/mol, in order to compute the molarity of the barium cation, we first need the moles in 7.17 g:

[tex]n_{Ba(CH_3COO)_2} = 7.17g*\frac{1molBa(CH_3COO)_2}{255.43g} =0.0281molBa(CH_3COO)_2[/tex]

Now, since one mole of barium acetate contains one mole of barium cations, we infer there are 0.0281 moles of barium cations. Moreover, since the molarity is computed by dividing the moles of those ones by the volume of the solution in liters, 350 mL (0.350 L) as it does not change, it turns out:

[tex]M=\frac{0.0281molBa^{2+}}{0.350L}\\\\M=0.0802M[/tex]

Best regards!

Answer:

Molecular solid

Explanation:

Molecular solids have a very low melting point. This is because, they are composed of weak intermolecular forces hence the layers of the solid easily fall apart as the solid melts.

We can see here that the solid being considered melts easily. Hence we can conclude that it is actually a molecular solid.

Answer:

Zinc

Explanation:

m = Mass of block [tex]1.8\times 10^{-2}\ \text{kg}=18\ \text{g}[/tex]

Dimension of block is 0.4981 inches by 0.531 inches, by 0.5839 inches

[tex]1\ \text{inch}^3=16.3871\ \text{mL}[/tex]

Volume of block

[tex]V=0.4981\times 0.531\times 0.5839\times 16.3871\ \text{mL}[/tex]

Density is given by

[tex]\rho=\dfrac{m}{V}\\\Rightarrow \rho=\dfrac{18}{0.4981\times 0.531\times 0.5839\times 16.3871}\\\Rightarrow \rho=7.112\ \text{g/mL}[/tex]

The density of the metal is [tex]7.112\ \text{g/mL}[/tex]. The metal here is zinc.

Answer:

The two elements are FLUORINE AND CHLORINE. These two elements have similar characteristics because they belong to the same group in the periodic table. On the periodic table, elements with similar properties are grouped together in the same group. Both chlorine and fluorine belongs to the halogen group.

Explanation:

Hope it helps, some how.

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

mg NaCl:

5g * (1000mg / 1g) = 5000mg

L Solution:

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

4.1 The Chemical Equation

LEARNING OBJECTIVES

Define chemical equation.

Identify the parts of a chemical equation.

A chemical reaction expresses a chemical change. For example, one chemical property of hydrogen is that it will react with oxygen to make water. We can write that as follows:

hydrogen reacts with oxygen to make water

We can represent this chemical change more succinctly as

hydrogen + oxygen → water

where the + sign means that the two substances interact chemically with each other and the → symbol implies that a chemical reaction takes place. But substances can also be represented by chemical formulas. Remembering that hydrogen and oxygen both exist as diatomic molecules, we can rewrite our chemical change as

H2 + O2 → H2O

This is an example of a chemical equation, which is a concise way of representing a chemical reaction. The initial substances are called reactants, and the final substances are called products.

Unfortunately, it is also an incomplete chemical equation. The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. If we count the number of hydrogen atoms in the reactants and products, we find two hydrogen atoms. But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.

What can we do? Can we change the subscripts in the formula for water so that it has two oxygen atoms in it? No; you cannot change the formulas of individual substances because the chemical formula for a given substance is characteristic of that substance. What you can do, however, is to change the number of molecules that react or are produced. We do this one element at a time, going from one side of the reaction to the other, changing the number of molecules of a substance until all elements have the same number of atoms on each side.

To accommodate the two oxygen atoms as reactants, let us assume that we have two water molecules as products:

H2 + O2 → 2H2O

The 2 in front of the formula for water is called a coefficient. Now there is the same number of oxygen atoms in the reactants as there are in the product. But in satisfying the need for the same number of oxygen atoms on both sides of the reaction, we have also changed the number of hydrogen atoms on the product side, so the number of hydrogen atoms is no longer equal. No problem—simply go back to the reactant side of the equation and add a coefficient in front of the H2. The coefficient that works is 2:

2H2 + O2 → 2H2O

There are now four hydrogen atoms in the reactants and also four atoms of hydrogen in the product. There are two oxygen atoms in the reactants and two atoms of oxygen in the product. The law of conservation of matter has been satisfied. When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is balanced. All proper chemical equations are balanced. If a substance does not have a coefficient written in front of it, it is assumed to be 1. Also, the convention is to use all whole numbers when balancing chemical equations. This sometimes makes us do a bit more “back and forth” work when balancing a chemical equation.

The total mass of insulin in the bottle : m = 4500 mg

Given

225 ml of insulin

The concentration 20 mg/ml

Required

The total mass of insulin

Solution

Density is the ratio of mass per unit volume

Can be formulated

ρ = m / V

m = ρ x V

Input the value :

m = 20 mg/ml x 225 ml

m = 4500 mg

Answer:

coefficients

Explanation:

Answer:

The correct answer is - one carbon atom and two oxygen atoms.

Explanation:

Carbon dioxide is a gas that is present in the atmosphere of the earth. The chemical formula of carbon dioxide is CO₂. In this colorless gas, there is one molecule of carbon dioxide is made up of one atom of carbon covalently double bonded to the two atoms of the oxygens.

Thus, the make- up of a carbon dioxide molecule (CO2) includes one carbon atom and two oxygen atoms.

whats the question?

Answer:

Chromuates are found in the chromosomes of the eukaryotic cells!!!!

Explanation:

I learned this in 7th grade!!!!

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Answer:

How the incident happened

Any chemicals involved in an incident

Any other hazards present in the lab

Explanation:

Above are the types of information that are necessary to communicate with emergency responders. The emergency responders ask the first question that how the incident happened. After that they ask that is there any harmful chemicals are present in the laboratory or what types of chemicals present in the laboratory. These questions were asked by the emergency responders in order to give the patient a suitable treatment.

The emergency responders are persons that helps in the case of emergency to deal with hazardous conditions. The information provided to the emergency responders helps in dealing with the condition.

The information that has to be provided to the emergency responder in case of a laboratory incident helps them to handle the situation. The questions about how the incident occurs, the presence of hazardous chemicals in the laboratory will help in avoiding the further hazard. The involvement of the chemical in the hazard will help to deal with the situation.

For more information about emergency responders, refer to the link:

https://brainly.com/question/20455351

Answer:

294.3 g/mol b.

The mass of 1.0*10^9 molecules of aspartame is 2.943*10^11 g/mol

ASPARTAME

Aspartame is  used as an artificial non-saccharide sweetener 200 times sweeter than sucrose, and is commonly used as a sugar substitute in foods and beverages.

CALCULATION

mass  of one molecule of aspartame = 294.3g/mol

mass of 1.0*10^9 molecules of aspartame =294.3*(1.0*10^9) g/mol

                                                                     =2.943*10^11 g/mol

refer https://brainly.com/question/25225559

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Answer:

Kp = 929551.4

Explanation:

First of all, we state the equilibrium:

4PBr₃ (g) ⇄ P₄(g) + 6Br₂(g)

In order to determine Kp, we need the partial pressure of each gas at equilibrium. Expression for Kp is:

{(Parial Pressure P₄) . (Partial Pressure Br₂)⁶} / (Partial Pressure PBr₃)⁴

Kp = 99.2 . 97.2⁶ / 97.4⁴

Kp = 929551.4

Take account that Kp can be also calculated from Kc.

Kp = Kc (RT)^Δn where Δn is the value for (final moles - initial moles) of any gas

Answer:

The electric force acting at a distance can be observed when someone rubs a glass rod with something woolen then the rod is taken away from the woolen material. The piece of paper leaps into air and sticks on to the rod. This is due to the electrostatic force of attraction that the paper sticks on the glass rod.

Explanation:

Answer:

Pam rubs a glass rod with wool, then moves the rod over a pile of small paper scraps. The scraps leap into the air and stick to the rod.

Answer:

CoSO4.7H2O

Cobalt II tetraoxosulphate VI heptahydrate

Explanation:

According to IUPAC nomenclature, compounds are named systematically.

We were told that there are seven water molecules in each formula unit Hence the correct formula of the compound is CoSO4.7H2O

According to IUPAC system, the metal is first named, followed by its oxidation state in Roman numerals. Then the name of the anion is mentioned as well as the number of water molecules. We have to take into account the correct prefix signifying the number of molecules of water of crystallization.

Answer:

d. none of the above (all have the same kinetic energy)

Explanation:

The kinetic theory of gases states that the molecules of an ideal gas experience a constant random motion.

At standard temperature and pressure (STP), the kinetic energy of an ideal gas such as hydrogen, argon, neon, sodium, oxygen, helium, magnesium, beryllium, nitrogen, carbon, fluorine, chlorine etc are all the same.

The standard temperature and pressure (STP) of an ideal gas is 273K and 100 kPa.

Hence, all of the gases have the same kinetic energy at standard temperature and pressure (STP).

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.