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3 years ago

In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq

2 answers:

5 0

Assuming that the reactants are:

(NH4)2SO4 (aq) + Ba(NO3)2 (aq)

and the products are:

BaSO4 (s) + 2NH4NO3 (aq),

then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.

According to the solubility rules, the following elements are considered insoluble when paired with SO4:

Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+

Therefore, the precipitate will be BaSO4 (s).

Mkey [24]3 years ago

5 0

Answer:

The precipitate is BaSO₄.

Explanation:

A precipitate is a chemical substance that in a liquid solution is in solid phase.

In the reaction:

(NH₄)₂SO₄(aq) + Ba(NO₃)₂(aq) → BaSO₄(s) + 2NH₄NO₃(aq)

(NH₄)₂SO₄, Ba(NO₃)₂ and NH₄NO₃ are in aqueous solution (aq) while BaSO₄ is in solid phase (s), that means that the precipitate is BaSO₄

I hope it helps!

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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.

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