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2 years ago

Why does a higher concentration make a reaction faster?

2 answers:

6 0

Option B is correct,
With increase in concentration the density of reactants increases and the system becomes more crowded, the greater the reactants will come in contact with each other and collisions occur. If collision is in proper orientation and has optimum energy then its fruitful and yields product. So, the greater the number of reactants, the greater will be the chances of collision and the greater will be the production of products per unit time and hence, greater is the rate of reaction.

Assoli18 [71]2 years ago

5 0

1 - B
2 - C
3 - A
4 -
5 - C
6 -
7 - C
8 - C
9 - B

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How many moles of calcium chloride are contained in a 333 gram sample?

Bogdan [553]

Answer & Explanation:

The molar mass of calcium chloride is 110.98 g/mol. We can use this information to solve this problem. We can set up our equation like this..

$\frac{333 g(CaCl2)}{} *\frac{1mol(CaCl2)}{110.98g(CaCl2)}$

Multiply straight across on the top and straight across on the bottom.

$\frac{333}{110.98}$

Now divide.

$\frac{333}{110.98}=3.00$

So, there are 3.00 moles of calcium chloride contained in a 33 gram sample which is answer choice D.

6 0

3 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

Which of the following represents a phase change? A. Melting B. Crushing C. Expanding D. Burning

Fantom [35]

Melting is an example of phase change

6 0

3 years ago

The energy exchange with the environment during a chemical reaction is called the _____ of reaction.

Allisa [31]

The temperature of reaction.

3 0

3 years ago

Read 2 more answers

How many moles are in 1.50x10^23 molecules of NH3

Rom4ik [11]

The answer is 0.25 moles because one mole of NH3 would have apporximately 6.02X10^23 molecules

6 0

3 years ago