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3 years ago

What would be the answer?

Chemistry

2 answers:

Anon25 [30]3 years ago

8 0

I believe it is C.....(don't take my word for it)

Alex Ar [27]3 years ago

3 0

The answer is b. The ball of paper, because it is more affected by gravity

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PLEASE I REALLY NEED ANSWER REAL QUICK

DENIUS [597]

Answer:

b. 760 g

Explanation:

The mass of the solution = 800 g

5% of NaCl by mass of the solution can be determined as follows;

5% of 800 = $\frac{5}{100}$ × 800

= 5 × 8

= 40 g

The mass of NaCl in the solution is 40 g.

The mass of water = mass of solution - mass of NaCl

= 800 - 40

= 760 g

Therefore, the mass of water required is 760 g.

5 0

3 years ago

A compound with the formula C6H12 may or may not be a saturated hydrocarbon.

astraxan [27]

A compound with the formula C6H12 is not considered a Saturated hydrocarbon.

Why is C6H12 isn't considered a Saturated hydrocarbon?

The ring's presence demonstrates that it is unsaturated. Keep in mind that the general formula for aliphatic hydrocarbons, CnH2n+2, serves as the foundation for its saturation. A chemical is unsaturated if it does not meet this requirement.

Example:

Hexane (C6H14)

C = 6; H = 14 = 2(6) + 2

resulting in hexane becoming saturated.

Cyclohexane(C6H12)

C = 6 and H = 12 do not equal 14 (x)!

cyclohexane is an unsaturated molecule as a result.

Cycloalkanes have the general formula C2H2n as well.

Hence, the given statement is false.

Learn more about the hydrocarbons here,

brainly.com/question/17578846

# SPJ4

6 0

1 year 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

At 338 mm hg and 72 c a sample of carbon monoxide gas occupies a volune of 0.225 L the gas transferred to a 1.50 L flask and the

Elis [28]

Answer:

P₂ = 0.09 atm

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 0.225 L

Initial pressure = 338 mmHg (338/760 =0.445 atm)

Initial temperature = 72 °C (72 +273 = 345 K)

Final temperature = -15°C (-15+273 = 258 K)

Final volume = 1.50 L

Final pressure = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂

P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L

P₂ = 25.83 atm .L. K / 293 K . L

P₂ = 0.09 atm

7 0

3 years ago

Which of the following is an example of a buoyant force acting on a piece of

hodyreva [135]

Answer:

D. The piece of bread travels with straight- line horizontal acceleration.

Explanation

4 0

3 years ago