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1 year ago

Does a negative ΔH mean that the heat should be treated as a reactant or as a product?

Chemistry

1 answer:

blagie [28]1 year ago

8 0

The negative ΔH mean that the heat should be treated as a reactant or as a product.

<h3>What is thermochemical equation? </h3>

The balanced equation which also include the enthalpy change of the reaction in it is termed as the thermochemical equation.

The change in enthalpy is denoted by ∆H . Depending on the process involved in the chemical equation, the value of ∆H can be negative or positive.

The reaction is said to have enthalpy value of change in enthalpy of a reaction. For instance, the value of ∆H is negative which implies that the reaction is exothermic.

This means that heat is released in the surrounding. Thus, the heat has to be entered in the product side.

Thus, we can say that negative ΔH mean that the heat should be treated as a reactant or as a product.

learn more about thermochemical equation:

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Magnesium oxide (MgO) forms when the metal burns in air. (a) If 1-25 9 of MgO contains 0.754 g of Mg, what is the mass ratio of

Vsevolod [243]

Answer:

Explanation:

a )

1.25 g MgO contains .754 g of Mg .Rest will be O

so oxide = 1.25 - .754 = 0.496 g

ratio of magnesium to oxide = .754/.496 = 1.52

b) 1.25 g of MgO contains .754 g of Mg

534 g of MgO contains .754 x 534 / 1.25 g = 322.11 g

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3 years ago

In which row has Aaron made an error in his chart?

hammer [34]

Answer:Row 2

Explanation:

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3 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

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3 years ago

A saline solution similar to that used for intravenous drips is made by dissolving 0.45 g sodium chloride in 50.00 g water. Whic

beks73 [17]

Answer:

E) 1, 2, and 3

Explanation:

50g H2O + 0.45g NaCl --> 50.45g saline solution

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3 years ago

A chemist prepares a solution by dissolving 0.3 mole of CaCl2 in enough water to make 1000 mL of solution. What is the molar con

Rasek [7]

$CaCl_{2}$ dissolves in water to give $Ca^{2+}$ and $Cl^{-}$ ions according to the following reaction:

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So, according to the above reaction, 1 mole of $CaCl_{2}$ produce 2 moles of $Cl^{-}$ ion,

So, 0.3 mole will give = 0.3 x 2 = 0.6 moles of $Cl^{-}$ ion

So, Molar concentration = $\frac{0.6moles}{1L} = 0.6$

Note: 1L = 1000mL

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3 years ago