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3 years ago

What is 25 °C in °F?

Chemistry

1 answer:

kifflom [539]3 years ago

7 0

Answer:

77°F

Explanation:

(25°C × 9/5) + 32 = 77°F

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How many grams are in 1 mole of Ar?

Ilia_Sergeevich [38]

Answer: 39.948 grams

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Ar, or 39.948 grams

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3 years ago

A(n) _______ is an organic compound in which a carbonyl group is bonded to a nitrogen atom. (2 points)

sineoko [7]

A(n )amide is an organic compound in which a carbonyl group is bonded to a nitrogen atom. This is <span>usually regarded as derivatives of carboxylic acids in which the hydroxyl group has been replaced by an amine or ammonia.</span>

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3 years ago

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An element has an atomic number of 27. How many protons and electrons are in a neutral atom of the element

Sergeu [11.5K]

Answer:

27 and 32

Explanation:

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2 years ago

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What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

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2 years ago

How many milligrams of sugar are in a teaspoon of sugar

grigory [225]

Answer 2325 milligrams.

Explanation:The unit is abbreviated as tsp. Convert Milligrams (mg) to Teaspoons (tsp): 1 mg is approximately equal to 0.0002 tsps. One milligram is a relatively small quantity of table sugar A mere one-teaspoon of granulated sugar contains some 2325 milligrams. Yay wiki

6 0

3 years ago