Ask question

Ask question

All categories

2 years ago

6

A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is 16.84 nm. What is the total wi

dth of a crystal in
millimeters 105 molecules across? Write your answer as a decimal.

1 answer:

nadezda [96]2 years ago

3 0

Answer:

u can just multiply 1

Explanation:

krkrfmfmmfdkd

You might be interested in

Which of the following is not a step in the conversion 1.3 Dam cubed to liters

lisabon 2012 [21]

A is the answer, and put the choices

7 0

3 years ago

How can we determine the formula for different molecular compounds

katen-ka-za [31]

Answer:

The first is the empirical formula which shows you the number of different atoms in the compound. After you convert the grams of each element into moles, you calculate the ratio of the moles, which gives you the ratio of the elements in the compound. More number-crunching gives you the molecular formula.

4 0

4 years ago

The combustion of ammonia in the presence of excess oxygen yields no2 and h2o: 4 nh3 (g) + 7 o2 (g) â 4 no2 (g) + 6 h2o (g) the

VLD [36.1K]

The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2
so 1mole of NH3 will give 1 mole of NO2
43.9 grams of NH3 contains 2.58 moles
so 2.58 moles will be produced of NO2
which is 118.7 grams this the amount of oxygen that is used.

8 0

3 years ago

The value of Ka for phenol (a weak acid), C6H5OH, is 1.00×10-10. Write the equation for the reaction that goes with this equilib

I am Lyosha [343]

Answer:

$Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Explanation:

Hello,

In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:

$C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+$

Thus, we can write the law of mass action, which consider the equilibrium concentrations of all the involved species, which is also known as the acid dissociation constant which accounts for the capacity the acid has to yield hydronium ions:

$K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Best regards.

3 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

3 years ago