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3 years ago

Plz, help me with this question!!!

Chemistry

1 answer:

zhenek [66]3 years ago

7 0

Answer:

There will be a change of odor, color, temperature, and composition. These are all the changes assosiated with a chemical change.

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Look at the reaction below. upper h subscript 2 upper s upper o subscript 4 (a q) plus upper c a (upper o upper h) subscript 2 (

kogti [31]

In the given reaction, the substance that is the base of the reaction is $CaSO_4$ . The correct option is C.

The base is a slippery liquid that is bitter and turns red litmus paper to blue.

Bases react with acids to form salts.

Then bases have pH above 7.

Examples of bases are calcium carbonate, sodium hydroxide, etc.

Thus, the correct option is C.

Learn more about bases

brainly.com/question/485375

8 0

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

4 years ago

Consider the following intermediate chemical equations.

mixas84 [53]

Answer:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

Explanation:

We have two equations:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

2H₂O(g) → 2H₂O(l)

To add the two equations: we omit H₂O(g) that is formed by 2 moles in the product side of the first equation and consumed by 2 moles from the reactants side in the second equation

So, the overall chemical equation is obtained by combining these intermediate equations is:

8 0

3 years ago

Read 2 more answers

What are three examples of fungi?

8_murik_8 [283]

Answer:

i would say the mushroom

Explanation:

i learned this before

3 0

3 years ago

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What is the volume of 400 g of metal with a density of 10 g/mL?

grigory [225]

Answer:

40 mL

Explanation:

Volume=400g/10(g/mL)

= 40mL

hope this will help you ❤️

4 0

3 years ago