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3 years ago

16 Which change results in the formation of different substances?(1) burning of propane

Chemistry

2 answers:

Fofino [41]3 years ago

8 0

Answer:The correct answer is option (1).

Explanation:

Burning of propane in the presence of oxygen will form carbon-dioxide, water and heat.The chemical reaction can represented as in chemical equation:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O+heat$

This is an example of chemical change where new product is formed.

Where as in melting of sold sodium chloride, deposition of carbon-dioxide gas and solidification of water ;phase of the substance is changing that is a physical change is taking place.

Hence, the correct option is (1).

Trava [24]3 years ago

6 0

1. The burning of propane creates different substances: propane and oxygen combine to create water and carbon dioxide:

C3H8 + 5 O2 becomes 3 CO2 + 4 H2O + heat

2. When NaCl melts, it is changing state, but it is still the same NaCl.

3. When gaseous CO2 deposits it is changing state, but it is still the same CO2.

4. When (presumably liquid) water solidifies, it is changing state, but it is still the same water.

So the answer is 1.

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

Write a paragraph

steposvetlana [31]

I think It’s 55 but that’s just me

6 0

3 years ago

To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?

Natalija [7]

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider $Mg(OH)_{2}$ as an insoluble substance, though it may be moderately soluble.

The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:

$Mg(NO_{3})_{2}_{(aq)}$ + $2NaOH_{(aq)}$ → $Mg(OH)_{2} {(s)}$ + $2NaNO_{3}_{(aq)}$

8 0

3 years ago

The names and chemical formulae of some chemical compounds are written in the first two columns of the table below. Each compoun

charle [14.2K]

Answer:

See below

Explanation:

Name Formula Major species

Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),

Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)

Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)

Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)

Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)

Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
The compounds containing metals are ionic. They produce ions in solution.
ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.

6 0

3 years ago

What stress will shift the following equilibrium system to the left?

Alona [7]

Answer: Increasing temperature

Explanation:

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a change of conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

The variation of one or several of the following factors can alter the equilibrium condition in a chemical reaction:

Temperature
The pressure
The volume
The concentration of reactants or products

In the case of the reaction in the question, the change that moves the balance to the left will be the one that moves it towards the reagents, that is, that favors the production of reagents instead of products.

Decreasing the concentration of SO3 and increasing the concentration of SO2 will favor the production of SO3, which is the product of the reaction.
Decreasing the volume increases the pressure of the system and the balance will move to where there is less number of moles. In the case of the reaction in question, we have 3 moles of molecules in the reactants (1 mole of O2 + 2 moles of SO2) while in the products there are 2 moles of SO3 only, therefore, decreasing the volume will displace the balance to the right, which corresponds to the sense in which there is less number of moles.

The reaction of the question is an exothermic since ΔH <0, therefore in the reaction heat is produced and it can be written in the following way,

2SO2(g) + O2(g) ⇌ 2SO3(g) + heat

So, if we increase the temperature we will be adding heat to the system, so the balance would move to the left to compensate for the excess heat in the system.

5 0

3 years ago