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3 years ago

What is the expected charge on an ion of nitrogen? A. -3 B. +3 C. +8 D. +5

Chemistry

1 answer:

Firlakuza [10]3 years ago

8 0

A. -3

hhhhhjhhhhhhhhhtghhhhhh (this is just bc the answer has to be at least 20 characters long don't mind this)

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It is estimated that 10 babies are born every 4 seconds. How many babies are born in 1 minute?

dlinn [17]

320 babies can be bor in a minute

6 0

3 years ago

The planet Mercury takes 176 days to do one spin around its pole. Compared to the Earth,

Tanya [424]

Answer:

The sun would appear to move more slowly across Mercury's sky.

Explanation:

This is because, the time it takes to do one spin or revolution on Mercury is 176 days (which is its period), whereas, the time it takes to do one spin or revolution on the Earth is 1 day.

Since the angular speed ω = 2π/T where T = period

So on Mercury, T' = 176days = 176 days × 24 hr/day × 60 min/hr × 60 s/min = ‭15,206,400‬ s

So, ω' = 2π/T'

= 2π/15,206,400‬ s

= 4.132 × 10⁻⁷ rad/s

So on Earth, T" = 1 day = 1 day × 24 hr/day × 60 min/hr × 60 s/min = ‭‭86,400‬ s

So, ω" = 2π/T"

= 2π/86,400‬ s

= 7.272 × 10⁻⁵ rad/s

Since ω' = 4.132 × 10⁻⁷ rad/s << ω" = 7.272 × 10⁻⁵ rad/s, the sun would appear to move more slowly across Mercury's sky.

4 0

2 years ago

A gas that exerts a pressure of 537 torr in a container with a volume of 5.30 L will exert a pressure of 255 torr when transferr

ArbitrLikvidat [17]

Answer:- The gas needs to be transferred to a container with a volume of 11.2 L.

Solution:- From Boyle's law. "At constant temperature, Volume is inversely proportional to the pressure."

It means, the volume is decreased if the pressure is increased and vice versa.

Here, the Pressure is decreasing from 537 torr to 255 torr. So, the volume must increase and calculated by using the equation:

$P_1V_1=P_2V_2$

Where, $P_1$ is initial pressure and $P_2$ is final pressure. Similarly, $V_1$ is initial volume and $V_2$ is final volume.

Let's plug in the values in the equation:

(537 torr)(5.30 L) = (255 torr)( $V_2$ )

$V_2=(\frac{537 torr*5.30 L}{255 torr})$

$V_2$ = 11.2 L

So, the new volume of the container needs to be 11.2 L.

4 0

3 years ago

Usually potassium hydrogen phthalate is kept very pure. But Stu Dent thinks the bottle of potassium hydrogen phthalate has been

Ahat [919]

Answer:

1.784 g

Explanation:

The equation of the reaction is;

NaOH(aq) + KHC8H4O4(aq) --------> KNaC8H4O4(aq) + H2O(l)

Number of moles of NaOH reacted = 17.47/1000 * 0.5000 M

Number of moles of NaOH reacted =8.735 * 10^-3 moles

From the reaction equation;

1 mole of NaOH reacted with 1 mole of KHC8H4O4

Hence, 8.735 * 10^-3 moles of NaOH reacts with 8.735 * 10^-3 moles of KHP.

So,

Mass of KHP reacted = 8.735 * 10^-3 moles * 204.2 g/mol = 1.784 g

5 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

3 years ago