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3 years ago

Will Mark brainliest fro correct answer!!!! Please answer in the picture chemistry

Chemistry

1 answer:

tiny-mole [99]3 years ago

6 0

Answer:

look at the graph

Explanation:

We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.

Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.

So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)

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anyanavicka [17]

0.066-0.0069= 0.0591

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2 years ago

To titrate 45.00 milliliters of an unknown HCl sample, use 25.00 milliliters of a standard 0.2000 M NaOH titrant. What is the mo

gtnhenbr [62]

V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L

M ( HCl ) = ?

V ( NaOH ) = 25.00 / 1000 => 0.025 L

M ( NaOH) = 0.2000 M

number of moles NaOH :

n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH

Mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH

moles HCl = 0.005 x 1 / 1

= 0.005 moles of HCl :

M ( HCl ) = n / V

M ( HCl ) = 0.005 / 0.045

= 0.1111 M

hope this helps!

4 0

3 years ago

What is the connection between diamonds and stardust?

Ede4ka [16]

According to an article dated back in February 8, 1992 which is entitled, “Science: Stardust is made of diamonds” on a website called newscientist (https://www.newscientist.com/article/mg13318073-000-science-stardust-is-made-of-diamonds/), American astronomers believed that diamonds are made in supernova explosions. It was said that the diamonds were the foundation of uncommon combinations of isotopes found in some meteorites. Donald Clayton of Clemson University in South Carolina suggested that the weightiest isotopes were more common in meteorites for the reason that the rare gases shaped in the neutron-rich outcome of a supernova explosion. Clayton also said, “the observed mixture of isotopes could have been produced only during the collapse of a massive star to form a neutron star”. This happens in a Type II explosion, for example the Supernova 1987A in the Large Magellanic Cloud. And rare gases like xenon become stuck in both weighty and light isotopes after the ejected gas from such a supernova cools down enough to create dust. The existence of the diamonds with these unusual gases in meteorites infers an alike source. Some of the carbon in the supernova fragments produces ordinary graphite dust, whereas some produces diamond dust. Considerable amount of stardust may be made of diamonds, if Clayton was not mistaken.

8 0

3 years ago

Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

Answer: The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

Explanation:

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

4 years ago

How do the ions in metals behave?

nikdorinn [45]

A metal ion is a type of atom compound that has an electric charge. Such atoms willingly lose electrons in order to build positive ions called cations.

4 0

4 years ago