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umka2103 [35]
3 years ago
12

Fill in the word that correctly completes this sentence.

Chemistry
1 answer:
crimeas [40]3 years ago
5 0
Mixtures are made up of 2 or more substances that are together in the same place but aren’t chemically bonded.
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To whoever helps me thank you so much have a wonderful day!
Schach [20]

Answer:

N and P

Explanation:

Anion:

When an atom gain the electrons anion is formed. The negative sign shows that atom gain electron because number of electron are greater than protons or we can say that negative charge becomes greater than positive charge.

Cation:

When atom lose electron cation is formed. The atom thus have positive charge because number of positive charge i.e protons are increased are greater than negative charge or electron.

In given problem N and phosphorus both can gain three electrons which means negative charge becomes greater that's why the extra electron gained by atoms are written as -3 and both form anion with charge -3.

while Al form cation with charge +3 Mg form cation with charge +2 and iodine and bromine both form anion with charge of -1.

8 0
3 years ago
4. Explain what you’ve learned about significant figures. How many significant figures are in the measurement 0.03050 kg?
nasty-shy [4]

There are 4 significant figures! Start counting after the first non-zero digit :)

Hope this helps.

3 0
3 years ago
As the temperature of a gas increases, which of the following explains the effect on the average kinetic energy of the gas?
lidiya [134]
As the temperature of a gas increases, the kinetic energy of the gas particles will also increase. As the temperature of the gas increase, the gas particles gains more energy to move faster, they thus collide more with one another and with the wall of the container, thus increasing pressure as well. So, as the temperature of a gas increases, the kinetic energy increases and the pressure increases as well if the gas is inside an inflexible container. 
8 0
3 years ago
Read 2 more answers
What is the scale that measures the acidity or basicity of a solution?
bearhunter [10]

Answer:

PH scale is a universal indicator of strongness of base and acid

3 0
3 years ago
Calculate Delta H in KJ for the following reactions using heats of formation:
lozanna [386]

Answer:

<h3>(a)</h3>

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

<h3>(b)</h3>

\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
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