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3 years ago

GIVING BRAINLIEST FIVE STARS AND 25 POINTS!

Chemistry

2 answers:

butalik [34]3 years ago

8 0

Answer:

There might be some problem in the engine's cooling system.

Explanation:

There are many reasons why your car could be overheating, but the basic process of overheating involves a problem with the vehicle's cooling system that doesn't allow heat to leave the engine compartment. This is often due to a leak or blockage somewhere in the engine cooling system or other components.

jasenka [17]3 years ago

3 0

Answer:

Engines can overheat for many reasons. In general, it's because something's wrong within the cooling system and heat isn't able to escape the engine compartment. The source of the issue could include a cooling system leak, faulty radiator fan, broken water pump, or clogged coolant hose.

Explanation:

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What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t

sukhopar [10]

The concentration of each ion in the solution of ammonium chloride is:

NH₄⁺ = 0.1886 M
Cl⁻ = 0.1886 M

To solve this problem, the formulas and the procedures that we have to use are:

M = n(solute)/v(solution) L
n = m / MW
MW= ∑ AWT

Where:

M= molarity
n = moles
m = mass
v = volume
MW = molecular weight
AWT = atomic weight

Information about the problem:

m(NH₄Cl) = 5 g
v(solution) = 500 ml
AWT (N)= 14 g/mol
AWT (H)= 1 g/mol
AWT (Cl) = 35 g/mol

Converting the volume units from (ml) to (L) we have:

v(solution) = 500 ml * (1 L/1000 ml)

v(solution) = 0,50 L

We calculate the moles of the NH₄Cl from the MW:

MW = ∑ AWT

MW (NH₄Cl)= AWT (N) + AWT (H)*4 + AWT (Cl)

MW (NH₄Cl)= 14 g/mol +( 1 g/mol * 4) + 35 g/mol

MW (NH₄Cl)= 14 g/mol + 4 g/mol + 35 g/mol

MW (NH₄Cl)= 53 g/mol

Having the MW we calculate the moles of NH₄Cl:

n(NH₄Cl) = m(NH₄Cl) / MW(NH₄Cl)

n(NH₄Cl) = m(H2SO4) / MW (H2SO4)

n(NH₄Cl) = 5 g / 53 g/mol

n(NH₄Cl) = 0.0943 mol

Applying the molarity formula, we get:

M(NH₄Cl) = n(NH₄Cl)/v(solution) L

M(NH₄Cl) = 0.0943 mol / 0,50 L

M(NH₄Cl) = 0.1886 M

There are 0.1886 moles of NH₄Cl per liter of solution.

Let's recognize that 1 mol NH₄Cl contains:

1 mol NH₄⁺
1 mol Cl⁻

The concentration of each ion is thus:

(1)*(0.1886 M) = 0.1886 M NH₄⁺

(1)*(0.1886 M) = 0.1886 M Cl⁻

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

Solvent
Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

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4 0

2 years ago

Can the excess reactant control the percentage yield? Please answer

Marina CMI [18]

Answer:

Yes, it can.

Explanation:

The percentage yield is decreased if the reactants do not completely form the products. To make a given mass of product, a process with a low percentage yield requires more of the reactants than a process with high percentage yield.

5 0

3 years ago

Read 2 more answers

Could someone help me asap?? (photo)

Gelneren [198K]

# 1- 50, #2-4,#3-a, #4-a

3 0

3 years ago

Four satellites are in orbit around the Earth. The heights and the masses of

tekilochka [14]

Answer:

Explanation:

4 0

3 years ago

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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

4 years ago