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Bad White [126]

3 years ago

14

Can anyone please help me with my chemistry homework? It’s about molarity. I can’t message anyone here because I’m new to brainl

y. But I will reply to you in the comment section and give you my information. So please help if you’re good at chemistry.

1 answer:

GREYUIT [131]3 years ago

4 0

Answer:

ok sure

Explanation:

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Which pair of elements form an ionic bond?

Anastasy [175]

Answer:

As an example I can say sodium (Na) and chlorine (Cl).

Explanation:

An ionic bond occurs when a metal element reacts with a nonmetal element. Therefore in the answer given above the Na is metal and Cl is nonmetal and they form a molecule through ionic bonding.

3 0

3 years ago

If you go from the Earth to the Moon, will your MASS change?

Solnce55 [7]

Answer:

No matter if you are on Earth, the moon or just chilling in space, your mass does not change. But your weight depends on the gravity force; you would weigh less on the moon than on Earth, and in space you would weigh almost nothing at all.

8 0

2 years ago

Read 2 more answers

Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given

professor190 [17]

Answer: -227 kJ

Explanation:

The balanced chemical reaction is,

$C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]$

$-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]$

$\Delta H_{C_2H_2}=-227kJ$

Therefore, the enthalpy change for $C_2H_2$ is -227 kJ.

7 0

3 years ago

PLZ HELP WITH THESE QUICK// WILL AWARD BRANLIEST

Tom [10]

Answer 1

The sample will dissolve in more than 1 minute.

Explanation :-

Generally Solubility and rate of solubility of substances increase with the increase in temperature. So lower the temperature less fast will it dissolve in the same amount of water.

Since at 50 C it takes 1 minute to dissolve, at 20 C which is lower temperature it will take more time to dissolve.

Answer 2

Transition metals

Explanation:-

Hardness depends on the extent of metallic bonding for metals. More the number of electrons more the metallic bonding.

Alkali metals with just 1 valence electron have weak metallic bonding. Alkaline earth metals have just 2 valence electrons. Transition elements has more electrons in penultimate shell and valence shell than lanthanides. Transition metals with most metallic bonding are the hardest

5 0

3 years ago

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas

n200080 [17]

Answer:

The limiting reactant is lead(II) nitrate.

7.20 g of precipitate are formed.

1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:

9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:

0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:

0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

5.76 - 3.86 = 1.9 g KCl

8 0

3 years ago