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3 years ago

Mushrooms, yeast, and mold are part of the Kingdom ____________.

Chemistry

1 answer:

Anit [1.1K]3 years ago

7 0

Answer:

fungi

Explanation:

because they are heterotrophs

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Ben is pushing a small trolley 10 N to the north and but Gary is pushing the trolley 5 N to the south. What is the total applied

andrew-mc [135]

Answer:

The total applied force on the trolley is 5 N toward the North.

Explanation:

In this problem, Ben is pushing the trolley to north and Gary is pushing it to south. So both the forces are acting 180° opposite to each other. As force is a vector quantity, the net force or total force acting on any object should be calculated by vector addition of number of forces along with their directions. So in this case, if we consider the force Ben is applying as F1 and the force Gary is applying as F2 on the trolley. Then the net or total force acting on the trolley will be $F1-F2$ . This is because, F1 and F2 are acting opposite to each other in direction.Thus, $Total force acting on the trolley = 10 N - 5 N$ .

So the total force acting on the trolley is 5 N and it is toward the north direction.

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4 years ago

What are the parts of the atom & which one is gained or lost in order for an object to

madam [21]

Answer:

The parts of an atom are electrons, usually found in an electron shell around the nucleus of the atom, and protons and neutrons, found in the nucleus of the atom. Typically, electrons are gained or lost in order to obtain charge.

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3 years ago

What is another name for the sugars organisms use for energy?1. Proteins2. Nucleic acids3. Carbohydrates4. Lipids

victus00 [196]

Answer:

Carbohydrates or Carbs have sugar molecules! :3 hope this helped you

Explanation:

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3 years ago

Which statement about electric and magnetic fields is true.

aksik [14]

The correct answer is 2. Magnetic fields must start at the north pole of a magnet and end at the south pole of a magnet. I took the test. Hope I helped!

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3 years ago

Read 2 more answers

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

6 0

3 years ago