need help with this question, anybody ..
-20 points :)

2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

When a solute is dissolved in a solvent, a solution is created. Dissolution is the process through which solutes, or dissolved parts, combine to form a solution inside a solvent. In this procedure, the gas, liquid, or solid dissolves inside the original solvent and forms a solution.

In some polymer applications, dissolution is also an issue since it results in swelling, a loss of strength and stiffness, and a change in volume. Whether a chemical process is man-made or natural, dissolution is crucial. Catalysts are tested using dissolution. 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

Therefore, 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

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Answer:

The bond between sodium sulfate is an ionic bond since it's a bond between a metal and non metals however the bond between sulfur and oxygen is a covalent bond since the two are non metals and the other reason that makes this an ionic bond is that there is both losing and gaining of electrons..

I hope this helps

Answer:

An alkaline is a substance that dissolves in water to produce hydroxyl ions (OH-)

Explanation:

The pH range of an alkaline is from 8–14.

Acidic pH ranges from 0–6.9.

Answer:

gu kha fuschhehdjdvdbeodbr

Answer:

D. Propanol

Explanation:

C3H7OH the presence of alcohol functional group makes it propanol

Answer:

all cells are produced from other preexisting cells through cell division

Answer:

2257000 J

Explanation:

Applying,

Q = Cₓm.................. Equation 1

Where Q = amount of energy need to boil the water, Cₓ = Enthalpy of vaporization of water, m = mass of water.

From the question,

Given: Cₓ = 2257000 J/kg, m = 1 kg

Substitute these values into equation 1

Q = 2257000×1

Q = 2257000 J

Hence the energy needed to boil all of the water is 2257000 J

Answer:

[ICl] = 0.0420 M

[I₂]  = [Cl₂] = 0.0140 M

Explanation:

Step 1: Calculate the initial concentration of ICl

[ICl] = 0.350 mol / 5.00 L = 0.0700 M

Step 2: Make an ICE chart

        2 ICl(g) ⇄ I₂(g) + Cl₂(g)

I        0.0700     0         0

C        -2x          +x        +x

E    0.0700-2x      x          x

The concentration equilibrium constant (Kc) is:

Kc = 0.110 = [I₂] [Cl₂] / [ICl]² = x² / (0.0700-2x)² = (x/0.0700-2x)²

0.332 = x/0.0700-2x

x = 0.0140

The concentrations at equilbrium are:

[ICl] = 0.0700-2x = 0.0700-0.0280 = 0.0420 M

[I₂]  = [Cl₂] = x = 0.0140 M

Answer:

NaCl, Na⁺,Cl⁻.

MgCl₂, Mg²⁺, Cl⁻.

CaO, Ca²⁺, O²⁻.

Li₃P, Li⁺, P³⁻.

Al₂S₃, Al³⁺, S²⁻.

Ca₃N₂, Ca²⁺, N³⁻.

FeCl₃, Fe³⁺, Cl⁻.

FeO, Fe²⁺, O²⁻.

Cu₂S, Cu⁺, S²⁻.

Cu₃N₂, Cu²⁺, N³⁻.

ZnO, Zn²⁺, O²⁻.

Ag₂S, Ag⁺, S²⁻.

K₂CO₃, K⁺, CO₃²⁻.

NaNO₃, Na⁺, NO₃⁻.

Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.

Al(OH)₃, Al³⁺,OH⁻.

Li₃PO₄, Li⁺, PO₄³⁻.

K₂SO₄, K⁺, SO₄²⁻.

Explanation:

Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.

Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.

Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.

Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.

Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.

Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.

Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.

Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.

Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.

Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.

Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.

Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.

Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.

Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.

Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.

Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.

Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.

Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.

Answer:

Sodium dihydrogenphosphate = NaH₂PO₄

Sodium monohydrogenphosphate = Na₂HPO₄

Explanation:

A buffer solution is a solution is a solution that resists changes to its oH when a little quantity of strong acid or strong base is added to it.

They are solutions of weak acids or weak bases and their salts known as conjugate base or conjugate acids respectively for the weak acids and weak bases.

For example, a solution of the weak acid ethanoic acid and its salt or conjugate base, sodium ethanoate serves as a buffer solution.

In biochemical experiments, where the pH of the reaction medium is kept as constant and as close as possible to that of the internal environment, buffer solutions are widely used. One of the commonly used buffers is the phosphate buffer. The phosphate buffer consists of the acid salts sodium dihydrogenphosphate and sodium monohydrogenphosphate. Sodium dihydrogenphosphate serves as the weak acid while sodium monohydrogenphosphate serves as the conjugate base.

The formulas of these two compounds are given below:

Sodium dihydrogenphosphate = NaH₂PO₄

Sodium monohydrogenphosphate = Na₂HPO₄

Answer:

Elements in the same period show trends in atomic radius, ionization energy, electron affinity, and electronegativity.

Answer:

Conversion of 2-methyl-2-butene into a secondary alkyl halide - ROOR, HBr

Conversion of 2-methyl-2-butene into a tertiary alkyl halide - HBr

Explanation:

The addition of HBr to 2-methyl-2-butene occurs in accordance to Markovnikov rule in the absence of peroxide.

According to Markovnikov rule; ''the negative part of the addendum is attached to the carbon atom bearing the least number of hydrogen atoms.'' Following the Markovnikov rule, the tertiary alkyl halide is obtained.

In the presence of peroxide, this rule is not followed and the reaction proceeds in an anti-Markovnikov way to yield a secondary alkyl halide.

Answer:

Hence the solids that should test to investigate the effects of cations on pH is

[tex]AlBr_{3}[/tex] (Cation is Al 3+)  

[tex]MgCl_{2}[/tex]  ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] ( Cation is NH2+).

Explanation:

The solids in the following should you test to investigate the effects of cations on pH.  

[tex]AlBr_{3}[/tex] contains (Cation is Al 3+)  

[tex]MgCl_{2}[/tex] contains ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] contains( Cation is NH2+ )

The atoms or the molecules containing the positive charge that gets attracted to the cathode are called cations. The compounds a. [tex]\rm AlBr_{3}[/tex], c. [tex]\rm MgCl_{2}[/tex] and e. [tex]\rm CH_{3}NH_{3}Br[/tex] should be investigated.

Cations are the positive charge containing molecules and atoms that have more protons in their nucleus than the number of electrons in their shells. They are formed when they lose one or more electrons to another atom.

The addition or release of the electrons of the cations and anions affects the pH system as absorption of the cation decreases the pH and absorption of the anions increases the pH.

Hence, [tex]\rm Al^{3+}[/tex], [tex]\rm Mg^{2+}[/tex] and [tex]\rm NH^{2+}[/tex] are the cation that should be investigated. The addition of the cations will reduce the pH of the reaction.

Therefore, absorption of the cation reduces the pH.

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Answer:

7 mol CuO

0.5 mol O₂

Explanation:

Step 1: Write the balanced equation

2 Cu + O₂ ⇒ 2 CuO

Step 2: Identify the limiting reactant

The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.

The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.

Since TMR > EMR, Cu is the limiting reactant

Step 3: Calculate the amount of CuO produced

7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO

Step 4: Calculate the excess of O₂ that remains

The amount of O₂ that reacts is:

7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂

The excess of O₂ that remains is:

4 mol - 3.5 mol = 0.5 mol

Answer:

False

Explanation:

Answer:

False

Explanation:

the effects are: temperature rises, water shortages, and increased fire threats

Answer: The mass of copper (II) nitrate produced is 105.04 g.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of copper = 35.5 g

Molar mass of copper = 63.5 g/mol

Plugging values in equation 1:

[tex]\text{Moles of copper}=\frac{35.5g}{63.5g/mol}=0.560 mol[/tex]

The given chemical equation follows:

[tex]3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)[/tex]

By the stoichiometry of the reaction:

If 3 moles of copper produces 3 moles of copper (II) nitrate

So, 0.560 moles of copper will produce = [tex]\frac{3}{3}\times 0.560=0.560mol[/tex] of copper (II) nitrate

Molar mass of copper (II) nitrate = 187.56 g/mol

Plugging values in equation 1:

[tex]\text{Mass of copper (II) nitrate}=(0.560mol\times 187.56g/mol)=105.04g[/tex]

Hence, the mass of copper (II) nitrate produced is 105.04 g.

The number of replaceable electrons in an atom is called its valency.

Examples

Thanks !

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Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

Answer:

3

Explanation:

The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.

As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Hence, M shell contains s,p and d subshells.

Answer:

[tex]m_{H_2O}=12.9gH_2O[/tex]

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the reaction whereby butane is combusted in the presence of excess oxygen:

[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

Thus, we can evidence a 2:10 mole ratio of butane to water, and thus, the stoichiometric setup to calculate the mass of produced water is:

[tex]m_{H_2O}=7.49gC_4H_{10}*\frac{1molC_4H_{10}}{52.12gC_4H_{10}} *\frac{10molH_2O}{2molC_4H_{10}}*\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=12.9gH_2O[/tex]

Regards!

Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.

Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.

A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.

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Answer:

89.04 g of NaNO₃.

Explanation:

We'll begin by converting 838 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

838 mL = 838 mL × 1 L / 1000 mL

838 mL = 0.838 L

Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:

Volume = 0.838 L

Molarity = 1.25 M

Mole of NaNO₃ =?

Mole = Molarity × volume

Mole of NaNO₃ = 1.25 × 0.838

Mole of NaNO₃ = 1.0475 mole

Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:

Mole of NaNO₃ = 1.0475 mole

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ =?

Mass = mole × molar mass

Mass of NaNO₃ = 1.0475 × 85

Mass of NaNO₃ = 89.04 g

Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

[tex]k=\frac{0.693}{t_1_/_2}[/tex]

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

[tex]k=\frac{0.693}{1.5 hrs}[/tex]

The expression for the rate constant is :

[tex]k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}[/tex]

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

[tex]\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6[/tex]

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

Answer:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.

Answer:

362g

Explanation:

heat lost by metal= heat gained by acetic acid

tfs are the same so you cando delta T

convert Cal/gc to J/gc

thectgod ig follow

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

Answer:

2.076

Explanation:

1 mole is 6.02 * 10^23

To convert from atoms (or molecules or compounds or ions etc.) to mols, you divide the number of atoms (or molecules or etc.) by 6.02 * 10^23

So it is (1.25 * 10^24)/(6.02 * 10^23)

=2.076

Answer:

[tex]\boxed {\boxed {\sf 2.08 \ mol \ C}}[/tex]

Explanation:

We are asked to convert a number of carbon atoms to moles.

We will use Avogadro's Number for this, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this problem, the particles are atoms of carbon. There are 6.022 ×10²³ atoms of carbon in 1 mole of carbon.

We will also use dimensional analysis to solve this problem. To do this, we use ratios. Set up a ratio using the underlined information.

[tex]\frac {6.022 \times 10^{23} \ atoms \ C}{1 \ mol \ C}[/tex]

We are converting 1.25 ×10²⁴ atoms of carbon to moles, so we multiply the ratio by that value.

[tex]1.25 \times 10^{24} \ atoms \ C* \frac {6.022 \times 10^{23} \ atoms \ C}{1 \ mol \ C}[/tex]

Flip the ratio. It remains equivalent, but it allows us to cancel the units atoms of carbon.

[tex]1.25 \times 10^{24} \ atoms \ C* \frac{1 \ mol \ C} {6.022 \times 10^{23} \ atoms \ C}[/tex]

[tex]1.25 \times 10^{24} * \frac{1 \ mol \ C} {6.022 \times 10^{23} }[/tex]

[tex]\frac{1.25 \times 10^{24} } {6.022 \times 10^{23} } \ mol \ C[/tex]

[tex]2.075722351 \ mol \ C[/tex]

The original measurement of atoms has three significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 5 in the thousandths place tells us to round the 7 up to an 8.

[tex]2.08 \ mol \ C[/tex]

1.25 ×10²⁴ atoms of carbon is equal to approximately 2.08 moles of carbon.

Answer:

[tex]{ \bf{HI _{(aq)} \: → \: H {}^{ + } _{(aq)} \: + \: \: I {}^{ - } _{(aq)} }}[/tex]

Answer: The mass of [tex]PH_3[/tex] produced is 45.22 g

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of [tex]P_4[/tex] = 62.0 g

Molar mass of [tex]P_4[/tex] = 124 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol[/tex]

Given mass of [tex]H_2[/tex] = 4.00 g

Molar mass of [tex]H_2[/tex] = 2 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol[/tex]

The chemical equation follows:

[tex]P_4(g)+6H_2(g)\rightarrow 4PH_3(g)[/tex]

By stoichiometry of the reaction:

If 6 moles of hydrogen gas reacts with 1 mole of [tex]P_4[/tex]

So, 2 moles of hydrogen gas will react with = [tex]\frac{1}{6}\times 2=0.333mol[/tex] of [tex]P_4[/tex]

As the given amount of [tex]P_4[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 6 moles of [tex]H_2[/tex] produces 4 mole of [tex]PH_3[/tex]

So, 2 moles of [tex]H_2[/tex] will produce = [tex]\frac{4}{6}\times 2=1.33mol[/tex] of [tex]PH_3[/tex]

We know, molar mass of [tex]PH_3[/tex] = 34 g/mol

Putting values in equation 1, we get:

[tex]\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g[/tex]

Hence, the mass of [tex]PH_3[/tex] produced is 45.22 g