Ask question

Ask question

All categories

4 years ago

13

If y gets smaller as x gets bigger, x and y have a(n) __ relationship.

1 answer:

Aleksandr [31]4 years ago

0 0

Answer:

Inverse relationship with each other

Explanation:

The inverse relationship cause increasing of one quantity and decreasing in the other

You might be interested in

34. Describe the spontaneous flow of heat between objects at different temperatures.

Temka [501]

A spontaneous flow of thermal energy takes place from hot to cold. In order to flow thermal energy from a colder object to a hotter object, the work has to be implied on the system. The particle at higher temperature exhibits more kinetic energy in comparison to the particle at a lower temperature. Thus, there is a continuous flow of heat between the objects from the higher temperature to the lower temperature.

5 0

3 years ago

3. What determines that an element is a metal?

Bingel [31]

Answer:

if it forms from positive ions and has types of metallic bonds

Explanation:

4 0

3 years ago

Read 2 more answers

Given the lattice energy of NaBr = 736 kJ/mol, the ionization energy of Na = 496 kJ/mol and the electron affinity of Br = −325 k

mars1129 [50]

Answer : The value of $\Delta H^o$ for the reaction is, -565 kJ/mol

Explanation :

The formation of sodium bromide is,

$Na(g)+Br(g)\overset{\Delta H^o}\rightarrow NaBr(s)$

$\Delta H^o$ = enthalpy of the reaction

The steps involved in the reaction are:

(1) Conversion of gaseous sodium atoms into gaseous sodium ions.

$Na(g)\overset{\Delta H_I}\rightarrow Na^{+1}(g)$

$\Delta H_I$ = ionization energy of sodium = 496 kJ/mol

(2) Conversion of gaseous bromine atoms into gaseous bromine ions.

$Br(g)\overset{\Delta H_E}\rightarrow Br^-(g)$

$\Delta H_E$ = electron affinity energy of bromine = -325 kJ/mol

(3) Conversion of gaseous cations and gaseous anion into solid sodium bromide.

$Na^+(g)+Br^-(g)\overset{\Delta H_L}\rightarrow NaBr(s)$

$\Delta H_L$ = lattice energy of sodium bromide = -736 kJ/mol

To calculate the $\Delta H^o$ for the reaction, the equation used will be:

$\Delta H^o=\Delta H_I+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$\Delta H^o=496kJ/mole+(-325kJ/mole)+(-736kJ/mole)$

$\Delta H^o=-565kJ/mole$

Therefore, the value of $\Delta H^o$ for the reaction is, -565 kJ/mol

7 0

3 years ago

Manganese(IV) sulfate<br> O Mn(SO4)2<br> O MnSO4<br> O MgSO4<br> DONE

oee [108]

Mn(SO4)2 is right opinion

7 0

3 years ago

Read 2 more answers

The concentration of H

Ivenika [448]

Answer:

Retype the question pls

Explanation:

??

3 0

3 years ago