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3 years ago

When 47.1 J of heat is added to 14.0 g of a liquid, its temperature rises by 1.80 ∘C. What is the heat capacity of the liquid?

Chemistry

1 answer:

Alja [10]3 years ago

3 0

Answer:

$\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}$

Explanation:

We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.

$q= mc\Delta T$

The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.

q= 47.1 J
m= 14.0 g
ΔT= 1.80 °C

Substitute these values into the formula.

$47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)$

Multiply the 2 numbers in parentheses on the right side of the equation.

$47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c$

$47.1 \ J = (25.2 \ g*\textdegree C) *c$

We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).

$\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}$

$\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c$

$1.869047619 \ J/g *\textdegree C = c$

The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.

$1.87 \ J/ g * \textdegree C =c$

The heat capacity of the liquid is approximately 1.87 J/g°C.

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How many moles of KBr are present in 500 ml of a 0.8 M KBr solution?

faltersainse [42]

Answer:

2) 0.4 mol

Explanation:

Step 1: Given data

Volume of the solution (V): 500 mL

Molar concentration of the solution (M): 0.8 M = 0.8 mol/L

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

500 mL × 1 L/1000 mL = 0.500 L

Step 3: Calculate the moles of KBr (solute)

The molarity is the quotient between the moles of solute (n) and the liters of solution.

M = n/V

n = M × V

n = 0.8 mol/L × 0.500 L = 0.4 mol

4 0

2 years ago

Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with

emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

It is a stichiometry problem.

2Na + 2H₂O → 2NaOH + H₂,

The balanced equation shows that 2.0 moles of Na metal react with 2.0 moles of water to produce 2.0 moles of NaOH and 1.0 mole of H₂,

Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: PV = nRT,

where, P is the pressure of the gas in atm (P at STP = 1.0 atm),

V is the volume of the gas in L (V = 7.80 L),

n is the number of moles in mole,

R is the general gas constant (R = 0.082 L.atm/mol),

T is the temperature of the gas in K (T at STP = 0.0 °C + 273 = 273.0 K).

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

Using cross multiplication:

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

6 0

3 years ago

I need an example of a chemical change involving water.Pleaseeee.

slava [35]

Chemical change occur when two substances are combined and produces a new substance or decomposes into two or more substances which are entirely different from the original two substances.

There are three types of chemical changes. These are 1) Inorganic Changes, 2) Organic Changes, and 3) Biochemical Changes

Here are some examples of chemicsal changes.

If you combine Sodium and Water, chemical changes causes decomposition into Sodium Hydroxide and Hydrogen.

Sodium + Water ==> Sodium Hydroxide and Hydrogen

Na + H2O ====> NaOH and H

Another example of chemical change is:

Carbon Dioxide and Water will decompose into Sugar and Oxygen

Carbon Dioxide + Water ==> Sugar and Oxygen

CO2 + H2O ==> CnH2nOn (where n is between 3 and 7) and O

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3 years ago

g For the reaction Ag2S(s)⇌2Ag+(aq)+S2−(aq)Ag2S(s)⇌2Ag+(aq)+S2−(aq), Keq=2.4×10−4Keq=2.4×10−4, and the equilibrium concentration

Roman55 [17]

Answer:

0.32 M

Explanation:

Step 1: Write the balanced reaction at equilibrium

Ag₂S(s) ⇌ 2 Ag⁺(aq) + S²⁻(aq)

Step 2: Calculate the concentration of Ag⁺ at equilibrium

We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Keq = [Ag⁺]² × [S²⁻]

[Ag⁺] = √{Keq / [S²⁻]}

[Ag⁺] = √{2.4 × 10⁻⁴ / 0.0023} = 0.32 M

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3 years ago

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ivolga24 [154]

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