What is the density of an object with a mass of 145.8g and an volume of 91.75 mL?

Answer:

3.83 mol/ L

Explanation:

Molarity is the amount of moles of a substance in a liter of solution.

1.8L ----- 6.9 mol of KCl

1L ----- 6.9 ÷1.8= 3.83 moles of KCl (3 s.f.)

Thus the molarity is 3.83 mol/L.

Answer: 3.8

Explanation:

Original answer is 3.83 but it needs significant figures so you need to round it to 2 significant figures

Answer:

8.0356 * 10^-5 moles of NaHCO3

Explanation:

Sulphuric acid = H2SO4

Sodium bicarbonate = NaHCO3

The reaction between both compounds is given by;

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

In the reactin above;

2 mol of NaHCO3 neutralizes 1 mol of H2SO4

At stp, 1 mol occupies 22.4 L;

1 mol = 22.4 L = 22400 mL

x mol = 0.9 mL

x = 0.9 / 22400 = 4.0178 * 10^-5 moles of H2SO4

Since 2 mol = 1 mol from the equation;

x mol = 4.0178 * 10^-5

x mol = 2 * 4.0178 * 10^-5

x = 8.0356 * 10^-5 moles of NaHCO3

Answer:enamel is formed by differentiated dental epithelial cells known as ameloblasts.

Explanation:

Answer:

Covalent compounds.

Explanation:

Hello,

In this case, when forming chemical bonds in order to form compounds, we say that if electrons are shared, covalent compounds are to be formed and they usually have subscripts that need prefixes to be named, for instance phosphorous pentachloride (PCl5), dichlorine heptoxide (Cl2O7), carbon tetrachloride (CCl4) and many others.

Regards.

Answer:

X 86 206

Explanation:

Radioactive atoms are nuclei that can under go disintegration to emit either an alpha particle, beta particle or gamma radiation. The process could be spontaneous or stimulated.

When a radioactive atom R 88 210 emits alpha particle, it would produce an element with atomic number 86 and mass number 206 i.e X 86 206. An alpha particle is usually a helium nucleus.

                     [tex]R^{210}[/tex] ⇒ [tex]x^{206}[/tex] + [tex]He^{4}[/tex] + energy

Answer:

c: is a good conductor of electricity

Explanation:

One of the physical properties of hydrochloric acid is that it is a good conductor of electricity in an aqueous state.

The dissolution of hydrochloric acid in water causes a complete dissociation into [tex]H^+[/tex] and [tex]Cl^-[/tex]. The free-floating ions that result from the dissolution enable hydrochloric acid to be able to conduct electricity.

In aqueous solution, the attraction of the positive dipole of the hydrochloric acid to the negative dipole of water and that of the positive dipole of water to the negative dipole of the hydrochloric acid creates a dipole-dipole interaction that allows complete dissociation of the hydrochloric acid. Thus, electrolyte results.

Correct option: c

Answer:

True

Explanation:

english please so i can help

Answer:

- [tex]n_{etOH}=0.11mol\ et OH[/tex]

- [tex]atoms_C=1.0x10^{23} atoms\ C[/tex]

Explanation:

Hello,

In this case, since ethanol has a molar mass of 46 g/mol, the moles in 5.0 g are:

[tex]n_{etOH}=5.0g\ etOH*\frac{1mol\ etOH}{46 g\ etOH} =0.11mol\ et OH[/tex]

Moreover, since the formula of glucose is C₆H₁₂O₆, its molar mass is 180 g/mol and six moles of carbon are in one mole of glucose (based on carbon's subscript), the atoms are computed by using the 6:1 mole ratio and the Avogadro's number as shown below:

[tex]atoms_C=5.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molC}{1molC_6H_{12}O_6} *\frac{6.022x10^{23}atoms\ C}{1molC} \\\\atoms_C=1.0x10^{23} atoms\ C[/tex]

Regards.

Answer:

A neutralization of a strong acid by a strong base gave out 5590 J of heat energy.

Determine the enthalpy change (in kJ) for the reaction.

Explanation:

A strong acid being neutralized by a strong base releases 5590 J of heat energy.

A chemical reaction in which an acid and a base interact quantitatively is known as neutralization or neutralisation. By neutralizing a reaction in water, surplus hydrogen or hydroxide ions are removed from the solution.

Neutralization is a sort of chemical reaction in which an acid and a base combine to produce salt and water. For instance: Reaction of sodium hydroxide and hydrochloric acid (HCl) Sodium hydroxide reacts with hydrochloric acid to form their salt sodium chloride (NaCl) and water.

To learn more about neutralization refer to:

https://brainly.com/question/23008798

#SPJ2

Answer:

9 protons

Explanation:

An element's atomic number is equal to the amount of protons in the nucleus - otherwise, the element's chemical identity would be different and it would be a completely different element.

Referencing any periodic table, we notice that fluorine has the atomic number of 9.

Because the atomic number is the same as the number of protons in the nucleus, fluorine has 9 protons.

Answer:

Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

Explanation:

The question requests to split the equation below into half equations;

Cl2+2Li⟶2LiCl

In redox chemistry, splitting into half equations simply means highlighting the reduction and oxidation reactions of the reaction.

Before proceeding, we hav to split the ionic compound; LiCl into it's component ions. So we have;

Cl₂ + 2Li ⟶ 2Li⁺Cl⁻

This leaves us with;

Cl₂ ⟶ 2Cl⁻ ............................... i

2Li ⟶ 2Li⁺  .............................. ii

Oxidation reactions can be identified by the increase in oxidation number and decrease in the case of reduction.

in reaction i, there is a decrease in oxidation number from 0 to -1. This is the reduction half equation,

in reaction ii, there is an increase in oxidation number from 0 to +1. This is the oxidation half equation

In terms of electrons, we have to even the charge;

Oxidation = Loss of electrons

Reduction = Gain of electrons

The half equations are given as;

Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

Separation of the redox reaction into its balanced component half-reactions is as follows;

Oxidation half-reaction;

Reduction half-reaction;

By definition;

Oxidation is simply characterized by an increase in oxidation number of a reacting entity.

Reduction is simply characterized by a decrease in oxidation number of a reacting entity.

A reaction in which oxidation and reduction occur simultaneously is termed a Redox reaction as in the given reaction.

Read more;

https://brainly.com/question/21405186

Answer:

d. 0.15 mol

Explanation:

Step 1: Given data

Step 2: Convert the mass of the solvent to kilograms

We will use the relationship 1 kg = 1,000 g.

66.6 g × (1 kg/1,000 g) = 0.0666 kg

Step 3: Calculate the moles of solute

The molality is equal to the moles of solute divided by the kilograms of solvent.

m = moles of solute / kilograms of solvent

moles of solute = m × kilograms of solvent

moles of solute = 2.4 mol/kg × 0.0666 kg

moles of solute = 0.15 mol

Answer:

Be

Explanation:

[tex]^94Be + ^42He \rightarrow ^{12}6C + ^10x[/tex]

Therefore, the element symbols is Be, which when bombarded by an α particle, will generate carbon-12 and a neutron.

Answer:

the answer is c

Explanation:

Answer:

1) If life exists in other planets

2) What type of living organisms exists on other planets

3) Predominant atoms/mixture/compound that exists on other planets

Explanation:

Earth and space science is the study of the earth, the space and other planets within the universe.

The following questions can be answered by studying earth and space science

1) If life exists in other planets

Over the years, there has been several reports suggesting this but a deep inquiry into this with scientific data can assist in putting the matter to rest.

2) What type of living organisms exists on other planets

If there is/are lives on other planets, what type of life/living organisms exist there. Are they plants? Are they animals? Or do we have a new classification for them?

3) Predominant atom/mixture/compound that exists on other planets

What is the most popular substance in each of these planets? The role of this substance in the survival of the planet or it's inhabitants also needs to be studied

Answer:

1. 33.43 g of KCl

2. 23.70 g of KBr

3. 50.45 g of Cr₂O₃

4. 18.82 g of SrO

Explanation:

Molar mass of  the elements and compounds in each of the reactions:

K = 39.0 g, Cl = 35.5 g, KCl = 74.5 g, Br = 80.0 g, KBr = 119.0 g, Cr = 52.0 g, O = 16.0 g, Cr₂O₃ = 152.0 g, Sr = 88.0 g, SrO = 104.0 g

1) 2K(s)+Cl2(g)/15.93G→2KCl(s)

From the mole ratio of the reaction above, 2 moles of K reacts with 1 mole of Cl₂ to give 2 moles of KCl

78.0 g (2 * 39.0 g) of K reacts with 71.0 g (2*35.5) of Cl₂ to produce 149.0 g(2*74.5) of KCl, therefore, Cl₂ is the limiting reactant.

15.93 g of Cl₂ will react to produce (149/71) * 15.93 of KCl = 33.43 g of KCl

2) 2K(s)+Br2(l)/15.93→2KBr(s)

From the mole ratio of the reaction, 2 moles of K reacts with 1 mole of Br₂ to give 2 moles of KBr

78.0 g (2 * 39.0 g) of K reacts with 160.0 g (2*80) of Br₂ to produce 238.0 g(2*119.0) of KBr, therefore, K is the limiting reactant which though is in excess.

15.93 g of Br₂ will react to produce (238/160) * 15.93 of KBr = 23.70 g of KBr

3) 4Cr(s)+3O2(g)/15.93→2Cr2O3(s)

From the mole ratio of the reaction, 4 moles of Cr reacts with 3 moles of O₂ to give 2 moles of Cr₂O₃

208.0 g (4 * 52.0 g) of Cr reacts with 96.0 g (3*2*16) of O₂ to produce 304.0 g (2*152.0) of Cr₂O₃, therefore, O₂ is the limiting reactant.

15.93 g of O₂ will react to produce (304/96) * 15.93 of Cr₂O₃ = 50.45 g of Cr₂O₃

4) 2Sr(s)/15.93+O2(g)→2SrO(s)

From the mole ratio of the reaction, 2 moles of Sr reacts with 1 mole of O₂ to give 2 moles of SrO

176.0 g (2* 88.0 g) of Sr reacts with 32.0 g (2*16) of O₂ to produce 208.0 g (2*104.0) of SrO, therefore, O₂ is the limiting reactant which though is in excess.

15.93 g of Sr will react to produce (208/176) * 15.93 of SrO = 18.82 g of SrO

Answer:

The answer is A

Explanation:

When nH2/nO2 =3/4 => it can cause explosion

Answer:

[tex]m_{MgCO_3}=0.0202molMgCO_3[/tex]

Explanation:

Hello,

In this case, for this purpose we first have to write the undergoing chemical reaction:

[tex]Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3[/tex]

Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:

[tex]n_{Na_2CO_3}=0.010L*0.300\frac{molNa_2CO_3}{1L}=0.003molNa_2CO_3 \\\\n_{Na_2CO_3}^{consumed}=0.006L*0.0400\frac{molMg(NO_3)_2}{1L}*\frac{1molNa_2CO_3}{1molMg(NO_3)_2} =0.00024molNa_2CO_3[/tex]

In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:

[tex]m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3} \\\\m_{MgCO_3}=0.0202molMgCO_3[/tex]

Regards.

Answer:

Final temperature = 100°C

Explanation:

The general equation of a coffee cup calorimeter is:

Q = m×C×ΔT

Where Q is the heat added to the system (3.6kJ = 3600J), m is the mass of the substance (200.0g), C is specific heat (0.240J /g°C) and ΔT is change in temperature).

Computing the values:

3600J = 200.0g×0.240J/(g°C)×ΔT

75°C = ΔT

ΔT is the difference in temperature between final and initial temperature. As intital temperature was 25°C:

75°C = Final T - 25°C

Answer:

Option C. 80.5%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Next, we shall determine the mass of C3H8 that reacted and the mass of CO2 produced from the balanced equation.

This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44 g/mol

Mass of C3H8 from the balanced equation = 1 x 44 = 44 g

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Mass of CO2 from the balanced equation = 3 x 44 = 132 g

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Next, we shall determine the theoretical yield of CO2.

This can be obtained as shown below:

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Therefore, 0.295 kg (i.e 295 g) will react to produce = (295 x 132)/44 = 885 g of CO2.

Therefore, the theoretical yield of CO2 is 885 g.

Finally, we shall determine the percentage yield of CO2 as follow:

Actual yield of CO2 = 712 g

Theoretical yield of CO2 = 885 g

Percentage yield of CO2 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield of CO2 = 712/885 x 100

Percentage yield = 80.5%

Therefore, the percentage yield of CO2 is 80.5%.

Answer:

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Explanation:

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Answer:

[tex]\large \boxed{\mathrm{subscripts}}[/tex]

Explanation:

Subscripts are numbers written after elements, subscripts indicate how many of each atom there are in a compound.

The atoms of each element in an compund is written in the subscript. The number of atoms present in any compound whether made up of only same element or different element is termed as Atomicity.

Given compound = [tex]\boxed{\sf{H_2O}}[/tex]

Here, there are two atoms of hydrogen and one atom of oxygen, So ratio of the atoms of elements in Water molecule = 2 : 1

━━━━━━━━━━━━━━━━━━━━

Answer:

Explanation:

Precision relates to instruments of measurement or scale . A scale measuring upto cm is less precise than an instrument measuring upto mm .

Here the first measurement appears to have been taken with an instrument measuring up to metre and the second measurement appears to have been taken with an instrument measuring up to decimeter . So second measurement is most precise .

Third measurement is closest to the actual measurement that is 45.0 m , so it is most accurate measurement .

Answer:

I believe the answer is a covalent bond

Answer:

6 mol/min

Explanation:

2A+2B→4C

The relationship between the reactants and products of this equation is given by;

1/2 -d[A]/dt = 1/2 -d[B]/dt = 1/4 d[C]/dt

Our focus is on A and C

From the question;

d[A]/dt = 3mol/min

We have;

1/2 (3) = 1/4 d[C]/dt

d[C]/dt = 4/2 * 3 = 6 mol/min

Answer:

The solvent is the substance which typically determines the physical state of the solution (solid, liquid or gas). The solute is the substance which is dissolved by the solvent.