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3 years ago

Mercury has a density of 13.56 g/ml. What is the volume in L of 52 kg of mercury?

Chemistry

1 answer:

sweet [91]3 years ago

5 0

Hello!

Given the density of mercury being 13.57 g/mL, and the mass of 52 kilograms, we need to find the volume.

To find the volume, we need to divide mass by density (V = m/d).

Notice that you are given 52 kilograms, but not grams. To convert kilograms to grams, you need to multiply it by 1000.

52 x 1000 = 52000 grams

With the correct measurements, we can find the volume.

V = 52000 grams / 13.56 grams/milliliter

V ≈ 3834.80826

Therefore, the volume of the mercury is about 3,834.81 mL.

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Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of potassium hydroxide and nickel

Gnom [1K]

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of nickel (II) acetate and potassium hydroxide is given as:

$Ni(CH_3COO)_2(aq.)+2KOH(aq.)\rightarrow 2CH_3COOK(aq.)+Ni(NO_3)_2(s)$

Ionic form of the above equation follows:

$Ni^{2+}(aq.)+2CH_3COO^-(aq.)+2K^+(aq.)+2OH^-(aq.)\rightarrow 2CH_3COO^-(aq.)+2K^+(aq.)+Ni(OH)_2(s)$

As, acetate and potassium ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ni^{2+}(aq.)+2OH^-(aq.)\rightarrow Ni(OH)_2(s)$

Hence, the net ionic equation is written above.

4 0

2 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?

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.

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