All categories

2 years ago

The temperature of a sample of liquid water changes from 50°C to 30°C. Which statement best explains the change that

Chemistry

1 answer:

Hunter-Best [27]2 years ago

8 0

Answer:

the molecules move more quickly and their average kinetic energy decreases :D

Explanation:

You might be interested in

Now can you answer pls

crimeas [40]

Answer:

cytoplasm is A, Dna is C, and nucleas is B

Explanation:

3 0

3 years ago

Which terms describe plastics that can and potentially be recycled?

hammer [34]

C describes plastics that can be recycled. Thermoplastic can be remolded and reheated after forming, whereas thermoset strengthens when heated

8 0

2 years ago

What do sodium and magnesium have in common?

lara [203]

Answer:

daddy will help

Explanation:

uhhhhhhhhhhhhhhhhhhh

7 0

2 years ago

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

4 0

3 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

6 0

3 years ago