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3 years ago

10. Which best explains why the total mass of the product(s) would be less than the total weight of the

Chemistry

1 answer:

Olegator [25]3 years ago

4 0

Answer:

D. Gases were released to the atmosphere

Explanation:

In accordance to the law of conservation of mass, the total amount of reactants must equate the total amount of products at the end of the reaction because matter can not be lost or created. However, certain changes like gas evolution, formation of precipitate etc. indicates the occurrence of a chemical reaction.

In a chemical reaction, the total mass of the product(s) would be less than the total weight of the reactant(s) because GASES, which constituted part of the mass of the reaction, WERE RELEASED INTO THE ATMOSPHERE. However, if the mass of the gas released can be accounted for, the amount of reactants and products must balance.

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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?

Alja [10]

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.

3 0

3 years ago

State the law of conservation of mass

Katen [24]

It’s basically that’s any system that’s closed to all transfers of matter and energy the mass of the system has to remain constant over time because they can’t change meaning you can’t add or remove from it

8 0

3 years ago

A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r

Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

5 0

2 years ago

Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at

MariettaO [177]

Answer:

P_2 =0.51 atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}$

$0.008389261745 \times 273 = 4.5P_2$

$P_2 =\dfrac{0.008389261745 \times 273 }{4.5}$

$P_2 =0.51 \ atm$

4 0

3 years ago

If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal

AVprozaik [17]

Answer:

The bismuth sample.

Explanation:

The specific heat $c$ of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g., $\rm 1\; ^{\circ}C$ .) The formula for specific heat is:

$\displaystyle c = \frac{Q}{m \cdot \Delta T}$ ,

where

$Q$ is the amount of heat supplied.
$m$ is the mass of the sample.
$\Delta T$ is the increase in temperature.

In this question, the value of $Q$ (amount of heat supplied to the metal) and $m$ (mass of the metal sample) are the same for all four metals. To find $\Delta T$ (change in temperature,) rearrange the equation:

$\displaystyle c \cdot \Delta T = \frac{Q}{m}$ ,

$\displaystyle \Delta T = \frac{Q}{c \cdot m}$ .

In other words, the change in temperature of the sample, $\Delta T$ can be expressed as a fraction. Additionally, the specific heat of sample, $c$ , is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.

Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth ( $\rm 0.123 \; J \cdot g\cdot \,^{\circ}C^{-1}$ .) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.

3 0

3 years ago