how many moles of H2O are needed to react with a mole of 3.0 of Na?​

whats the question?

Answer:

convection is not form of energy

Answer:

(240 × 3 × 31.998) ÷ (122.5 × 2) grams

Explanation:

To solve this problem, it is best to use the mole concept approach. Here find the number of moles of the given specie and use it to find that of the unknown.

Then convert the number of moles of the unknown to mass.

     Number of moles  = [tex]\frac{mass}{molar mass}[/tex]  

  Mass = number of moles x molar mass

Mass of potassium chlorate = 240g

The balanced reaction equation is given as:

              2KClO₃  →  2KCl  +  3O₂

Number of moles of KClO₃  = [tex]\frac{240}{122.5}[/tex]  

     Mole ratio:

              2 mole of KClO₃  will produce 3 moles of O₂

            [tex]\frac{240}{122.5}[/tex]  mole of KClO₃ will produce  [tex]\frac{240}{122.5}[/tex] x 3 / 2 mole of O₂

So, mass of oxygen;

        Mass  =   [tex]\frac{240}{122.5}[/tex] x 3 / 2   x 31.998  =   [tex]\frac{240}{122.5}[/tex]  x 3 x [tex]\frac{1}{2}[/tex]  x 31.998

 This gives (240 × 3 × 31.998) ÷ (122.5 × 2) grams

In the experiment the potassium and chlorate decompose as per the chemical equation of KClO3 → KCl + O2.

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Answer:

A

Explanation:

Correct on edge

The temperature and moles of gas constants would most likely be filled in the first row of the table. Therefore, option A is correct.

The gas constant, denoted by R, is a physical constant that is expressed in terms of units of energy per temperature increment per mole. It's also referred to as the ideal gas constant, the molar gas constant, or the universal gas constant.

A general constant in the equation of state of gases that is equal to the product of the pressure and volume of one mole divided by the absolute temperature in the case of an ideal gas see gas law sense c.

All gases adapt an equation of state known as the ideal gas law: PV = nRT, where n is the number of moles of the gas and R is the universal gas constant.

Thus, option A is correct.

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Your question is incomplete, most probably your question was

The incomplete table below shows selected characteristics of gas laws.

Name

Variables

Constants

Equation

?

?

?

PV = k

Charles's law

?

?

V = kT

?

Temperature, pressure

?

P = kT

Combined gas law

Pressure, temperature, volume

?

?

Which constants would most likely be filled in the first row of the table?

temperature and moles of gas

volume and temperature

pressure and volume

pressure and moles of gas

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:

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:

a.  x = + 2

b.  x = + 3

c.  x = + 2

d.  x = + 3

Explanation:

The oxidation number is a formal charge assigned to an atom present in a molecule or formula unit or ion based on some arbitrary rules.

a.

HgO

The oxidation number of Hg in HgO is:

x + (-2) = 0

x = +2

b.

Al₄C₃

The oxidation number of Al in Al₄C₃ is:

4x +(3 × -4) = 0

4x - 12 = 0

4x = +12

x = 12/4

x = +3

c. CrF₂

x + ( 2 × - 1) = 0

x - 2 = 0

x = + 2

d. Fe₂S₃

2x + ( 3 × - 2) = 0

2x + (-6) = 0

2x = 6

x = 6/2

x = +3

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

The  formula of the hydrate = CuSO₄• 3H₂O

Given

4.175 grams sample CuSO₄• xH₂O

3.120 grams anhydrous compound CuSO₄

Required

The formula

Solution

mass of H₂O driven off :

= 4.175 - 3.12

= 1.055 g

MW CuSO₄ = 159.5 g/mol

MW H₂O = 18 g/mol

mol ratio of CuSO₄ : H₂O :

= 3.12/159.5 : 1.055/18

= 0.01956 : 0.05861

= 1 : 3

Based on the data provided, the formula of the hydrated salt is CuSO4.3H20

The formula of the hydrate is determined from the mole ratio of the anhydrous saltand water.

Mass of water = 4.175 - 3.120

mass of water = 1.055 g

Mole ratio= mass/molar mass

molar mass of CuSO4 = 160 g/mol

molar mass of H2O = 18 g/mol

CuSO4 = 3.120/160 = 0.0195

H2O = 1.055/18 = 0.0586

CuSO4 = 0.0195/0.0195 = 1

H2O = 0.0586/0.0195 = 3

Therefore, the formula of the hydrated salt is CuSO4.3H20

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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:

I think it will decrease

The water solution of KCl is the BEST conductor of electricity (Option B).

In conclusion, the water solution of KCl is the BEST conductor of electricity (Option B).

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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:

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:

The number of valence electrons increases from top to bottom within each group.

Explanation:

The valence electrons are the outermost shell electrons.

Therefore, the number of valence electrons does not increase from top to bottom within each of the group.

Down a group, the number of valence electrons remains the same.

For group I elements, the number of electrons is 1.

 The group is the vertical arrangement of elements.

From left to right across a period, the number of valence electrons increases because you move from group to group.

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.

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Answer:

When titanium (II) nitride forms an ionic compound, TiN is the chemical formula of the compound.

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:

n = 3.25 mol

Explanation:

Given data:

Pressure of sample = 2 atm

Volume of sample = 40 L

Temperature = 300 K

Number of moles = ?

Solution:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

2 atm ×40 L = n × 0.0821 atm.L/mol.K × 300 K

80 atm.L = n  × 24.63 atm.L/mol

n = 80 atm.L  /24.63 atm.L/mol

n = 3.25 mol

Answer:

a. BaCl2 is ionic compound.

b. CH4 is covalent compound.

b. MnO2 is ionic compound.

e. H2S is covalent compound.

c. H3AsO4 is ionic compound.

f. CH3OH is covalent compound.

Explanation:

BaCl2, MnO2 and H3AsO4 are ionic compounds due to the presence of ionic bond between the atoms while on the other hand, CH4, H2S and CH3OH are covalent compounds due to the presence of covalent bond between their atoms. In covalent bonds, sharing of electrons occur whereas in ionic bonds, complete transfer of electrons occur from one atom to another atom.

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

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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.

Answer:

The object is moving at a constant speed.

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

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.