All categories

2 years ago

The particles in both gases and liquids select one:

1 answer:

6 0

A is the answer, it cant be B cause liquid cant change with ither particles. It cant be C cause gases dont vibrate they slide over each ither. It cant be D cause solid is the one closely packed together

You might be interested in

What charge would you expect for an formed by S?

sp2606 [1]

Answer:

in this the correct answer is option 2.

6 0

2 years ago

What is the Kb expression for aniline (C6H5NH2)?

Dimas [21]

This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]

hope this helps:)

</span>

4 0

3 years ago

Read 2 more answers

A Snickers bar contains 27.0 g of sugars (carbohydrates), 4.0 g of protein and 12.0 g of fat. How many food calories (Cal) are c

amm1812

Answer:

There are 232 calories in the bar

Explanation:

Carbohydrates and protein are both 4 calories per gram, while fat is 9 calories per gram.

$27*4=108\\4*4=16\\12*9=108\\108+108+16=232$

4 0

2 years ago

Which of the following factors does NOT affect surface currents?

lubasha [3.4K]

Answer: B

Explaining: surface currents are caused by wind density affects deep ocean currents

4 0

2 years ago

Read 2 more answers

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

2 years ago