What is the name of this compound

Which of the following keeps the planets in their orbits in the Solar System?

posledela

Answer:

B- gravity

Explanation:

Gravity is a force in which everything in the universe is trying to pull everyother thing towards themsef. Here, the sun's gravity keeps the planet in orbit.

I hope im right! !!

6 0

2 years ago

Which of these have the same number of particles as 1 mole of water H2O

Fed [463]

Answer:

It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.

Explanation:

The question is not very much clear.

If you are asking for molecules then 1 mole water= 6.023 * 10^23

If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3

If you are asking for particles then,

So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.

I hope that was helpful!

H=1 proton,1 electron

O=8 protons,8 neutrons and 8 electrons

total particles in one H2O molecule-28

total no. of particles in 1 mole of water- 6.023 * 10^23 * 28

8 0

2 years ago

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. <u>Therefore, the given reaction is exothermic.</u>

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

<u>Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.</u>

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. <u>Therefore, the given combustion reaction is an oxidation-reduction reaction.</u>

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

<u>Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.</u>

3 0

2 years ago

7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.

beks73 [17]

Answer:

$\Delta H_{f,C_3H_4}=276.8kJ/mol$

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

$\Delta H_{rxn} =- m_wC_w\Delta T$

We plug in the mass of water, temperature change and specific heat to obtain:

$\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ$

Now, this enthalpy of reaction corresponds to the combustion of propyne:

$C_3H_4+4O_2\rightarrow 3CO_2+2H_2O$

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

$\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}$

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

$\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol$

Now, we solve for the enthalpy of formation of C3H4 as shown below:

$\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}$

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

$\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol$

Best regards!

7 0

2 years ago

What is a common use of bases?

Fittoniya [83]

A common use of bases is to reduce indigestion. Hence, option D is correct.

A base is a substance that can neutralize the acid by reacting with hydrogen ions.

Bases are substances whose solutions have a pH value of more than 7. Bases are bitter in taste.

Whereas acids are the substances whose solutions have a pH value of less than 7. Acids are sour in taste.

When an excess of acid is formed in the human stomach then an antacid tablet which is basically a base is consumed that helps in reducing indigestion.

Therefore, we can conclude that the common use of bases is to reduce indigestion.

Learn more about the base here:

brainly.com/question/1432645

#SPJ1

3 0

1 year ago