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3 years ago

So, a gas, and an

Chemistry

1 answer:

valina [46]3 years ago

5 0

Answer:

d) cut the large sized Cu solid into smaller sized pieces

Explanation:

The aim of the question is to select the right condition for that would increases the rate of the reaction.

a) use a large sized piece of the solid Cu

This option is wrong. Reducing the surface area decreases the reaction rate.

b) lower the initial temperature below 25 °C for the liquid reactant, HNO3

Hugher temperatures leads to faster reactions hence this option is wrong.

c) use a 0.5 M HNO3 instead of 2.0 M HNO3

Higher concentration leads to increased rate of reaction. Hence this option is wrong.

d) cut the large sized Cu solid into smaller sized pieces

This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.

You might be interested in

What is the mass in grams of 1.50 X 1023 molecules of NaOH?

avanturin [10]

Answer:

a. 10.0 g

Explanation:

The molar mass of NaOH is 39.997 grams which was calculated by adding the molar mass of Sodium, Oxygen, and Hydrogen $22.990 + 15.999 + 1.008 = 39.997$ . But before we do anything with this number we must first convert $1.50 * 10^{23}$ molecules to moles. We can do this by dividing by Avogadro's number or $6.022*10^{23}$ ( $\frac{1.50*10^{23}}{6.022*10^{23}} = 0.24909$ moles). We then convert moles to grams by multiplying by the molar mass of NaOH, ( $0.24909 * 39.997 = 9.96$ grams). This is almost exactly 10 and the margin of error is likely due to the original numbers not being precise enough.

5 0

4 years ago

A lactic acid/lactate ion buffer solution contains 0.14 M HC3H5O3 and 0.83 M C3H5O3−, respectively. The Ka value of lactic acid

yKpoI14uk [10]

Use Henderson Hasselbache
PH=pka+log([salt]/[acid])
Pka=-log(ka)
-log(1.4x10^-4)= 3.85
log(.83/.14)= .77
3.85+.77= 4.62
PH=4.62

6 0

3 years ago

Read 2 more answers

What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?

aivan3 [116]

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

3 0

3 years ago

How many moles are in 38 grams of Argon? Round to 2 decimal places. Use "e" for exponents

yuradex [85]

Answer: 39.948

Explanation:

How many grams Argon in 1 mol? The answer is 39.948. We assume you are converting between grams Argon and mole. You can view more details on each measurement unit: molecular weight of Argon or mol The molecular formula for Argon is Ar.

3 0

3 years ago

Look at the following data provided below:

Vlad1618 [11]

Considering the Hess's Law, the enthalpy change for the reaction is -84.4 kJ.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>Enthalpy change for the reaction in this case</h3>

In this case you want to calculate the enthalpy change of:

2 C (graphite) + 3 H₂(g) → C₂H₆(g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: C₂H₆(g) + $\frac{7}{2}$ O₂(g) → 2 CO₂(g) + 3 H₂O(l) ; ΔH° = –1560 kJ

Equation 2: H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l) ; ΔH° = –285.8 kJ

Equation 3: C(graphite) + O₂(g) → CO₂(g) ; ΔH° = –393.5 kJ

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 2 moles of C(graphite) on reactant side and it is present in third equation. In this case it is necessary to multiply it by 2 to obtain the necessary amount. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

Now, you need 3 moles of H₂(g) on reactant side and it is present in second equation. In this case it is necessary to multiply it by 3 to obtain the necessary amount and the variation of enthalpy also is multiplied by 3.

Finally, 1 mole of C₂H₆(g) must be a product and is present in the first equation. Since this equation has 1 mole of C₂H₆(g) on the reactant side, it is necessary to locate the C₂H₆(g) on the reactant side (invert it). When an equation is inverted, the sign of delta H also changes.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 CO₂(g) + 3 H₂O(l) → C₂H₆(g) + $\frac{7}{2}$ O₂(g); ΔH° = 1560 kJ

Equation 2: 3 H₂(g) + $\frac{3}{2}$ O₂(g) → 3 H₂O(l) ; ΔH° = –857.4 kJ