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3 years ago

Do you think sodium carbonate is an acid or base?

1 answer:

7 0

I just looked it up and it says "Na2CO3 (sodium carbonate) is neither an acid nor a base. It is a salt..." I really hoped this helped you, if not I'm sorry.

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12) A 17.92 g sample of copper was submerged in water, and the volume reading rose by 2.00 mL. What is the density of the copper

emmasim [6.3K]

Hey there!:

Mass = 17.92 g

Volume = 2.00 mL

Density = ??

Therefore:

D = m / V

D = 17.92 / 2.00

D = 8.96 g/mL

Hope this helps!

3 0

3 years ago

For the following reactions, predict the products and write the balanced formula equation, complete ionic equation, and net ioni

stealth61 [152]

Answer:

Explanation:

To predict the products of these reactions we need to know the kind of reactions. All these reactions are double replacement reaction. In these kinds of reactions, the products will be the result of exchanging ions in the reactants. So, the first step is to identify the ions.

For the reaction, we have Hg2(NO3)2 and CuSO4. We have the ions Hg+1, NO3-1, Cu+2 and SO4-2

The way to make this exchange is putting together positive in one species with the negative of the other species. Following that rule we have

$Hg^{+1} - - - (SO_{4})^{-2}[/text]
the oxidation number will tell you the subscript for each species in the compound. In this case, is Hg2(SO4) [tex]Cu^{+2} - - - (NO_{3})^{-1} - - -> Cu(NO_{3})_{2} [/text] 
So, the products for this reaction will be
 [tex]Hg_{2} (NO_{3})_{2}(aq) + CuSO_{4}(aq) --> Hg_{2}SO_{4} + Cu(NO_{3})_{2}[/text]

After this, we proceed to balance the equation. For this, we check that we have the same number of each element on both sides of the equation. In this case, we can see that we have the same number, so the equation is balanced. Finally, we check the rules of solubility to see if the species are soluble in water or not. In this case sulfates area always soluble except for mercury so Hg2(SO4) precipitates in the solution (pre). Nitrates are always soluble so Cu(NO3)2 is soluble (aq) 
[tex] Hg_{2}(NO_{3})_{2}(aq) + CuSO_{4}(aq) - -> Hg_{2}SO_{4} (pre) + Cu(NO_{3})_{2}(aq)$

The complete ionic equation allows to show which of the reactants or products exist primarily as ions. For this reaction this will be:

$2Hg^{+1}(aq) + 2(NO_{3})^{-1}(aq) + (SO_{4})^{-2}(aq) + Cu^{+2}(aq) --> Hg_{2}SO_{4} (pre)+ Cu^{+2}(aq) + (NO_{3})^{-1}(aq) [/text]

To get net ionic equation we take away the ions that did not participate in the reactions. In other words the ones that are the same on both sides in the equation. In this case we see that [tex] Cu^{+2}(aq) and (NO_{3})^{-1}(aq) [/text] are the same on both sides so those ions are not include in the net ionic equation. This is:
[tex] 2Hg^{+1}(aq) + (SO_{4})^{-2}(aq) --> Hg_{2}SO_{4} (pre) [/text]

B [tex] Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq)$

ions (1) $Ni^{+2} and (NO_{3})^{-1}$

ions (2) $Ca^{+2} and Cl^{-1}$

Exchanging

$Ni^{+2} ---- Cl^{-1} --> NiCl_{2}$

$Ca^{+2} --- (NO_{3})^{-1} --> Ca(NO_{3})_{2}$

Products

$Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq) --> NiCl_{2} + Ca(NO_{3})_{2}$

The equation is already balanced

Chlorides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NiCl2 is soluble (aq)

Nitrates are always soluble. Ca(NO3)2 is soluble (aq)

Since both compounds are soluble, we can say that there is not reaction.

Complete ionic equation

$Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq) - - > Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq)$

Net ionic equation:

The ions in both sides of the equation are the same so all of them are cancelled and we cannot get a net ionic equation this explains why there is no reaction in this case.

C $K_{2}CO_{3}(aq) + MgI_{2}(aq)$

Ions(1) $K^{+1} and (CO_{3})^{-2}$

Ions(2) $Mg^{+2} and l^{-1}$

Exchanging

$K^{+1} --- l^{-1} - - > KI$

$Mg^{+2} --- (CO_{3})^{-2} - - > Ca(CO_{3})$

Products

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> Kl + MgCO_{3}$

The equation is not balanced

Balance equation is

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> 2Kl (aq) + MgCO_{3} (pre)$

iodides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. KI is soluble (aq)

carbonates are always insoluble except group 1 cations. MgCO3 is insoluble (pre)

complete ionic equation

$2K^{+1}(aq) + (CO_{3})^{-2}(aq) + Mg^{+2}(aq) + 2l^{-1}(aq) - - > MgCO_{3} (pre) + 2K^{+1}(aq) + 2l^{-1}(aq)$

Net ionic equation

$(CO_{3})^{-2}(aq) + Mg^{+2}(aq) - - > MgCO_{3} (pre)$

D $Na_{2}CrO_{4}(aq) + AlBr_{3}(aq)$

Ions(1) $Na^{+1} and (CrO_{4})^{-2}$

Ions(2) $Al^{+3} and Br^{-1}$

Exchanging

$Na^{+1} ---- Br^{-1} - -> NaBr$

$Al^{+3} --- (CrO_{4})^{-2} - -> Al_{2}(CrO_{4})_{3}$

Products

$Na_{2}CrO_{4}(aq) + AlBr_{3}(aq) - -> NaBr + Al_{2}(CrO_{4})_{3}$

The equation is not balanced

Balance equation is

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr + Al_{2}(CrO_{4})_{3}$

bromides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NaBr is soluble (aq)

chromates are always insoluble except group 1 cations. Al2(CrO4)3 is insoluble (pre)

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr(aq) + Al_{2}(CrO_{4})_{3}(pre)$

Complete ionic equation

$6Na^{+1}(aq) + 3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) + 6Br^{-1}(aq) - -> Al_{2}(CrO_{4})_{3}(pre) +6Br^{-1}(aq) + 6Na^{+1}(aq)$

Net ionic equation

$3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) - -> Al_{2}(CrO_{4})_{3}(pre)$

6 0

3 years ago

The diagram shows the movement of particles from one end of the container to the opposite end of the container.

Aleksandr-060686 [28]

The correct option is this: EFFUSION BECAUSE THERE IS A MOVEMENT OF A GAS THROUGH A SMALL OPENING INTO A LARGER VOLUME.
Effusion refers to the movement of gas particles through a small hole. According to Graham's law, the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.

7 0

3 years ago

Read 2 more answers

A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.

Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0

3 years ago

If a 100. -g sample of a hydrated compound contains 37.07-g sodium, 48.39-g carbonate and 14.54-g water, find the empirical form

Mumz [18]

he required empirical formula based on the data provided is Na2CO3.H2O.

<h3>What is empirical formula?</h3>

The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.

We have the following;

Mass of sodium = 37.07-g

Mass of carbonate = 48.39 g

Mass of water = 14.54-g

Number of moles of sodium = 37.07-g/23 g/mol = 2 moles

Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole

Number of moles of water = 14.54/18 g/mol = 1 mole

The mole ratio is 2 : 1: 1

Hence, the required empirical formula is Na2CO3.H2O

Learn more about empirical formula : brainly.com/question/11588623

3 0

1 year ago