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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

The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi

weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%$

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%$

So the answer to the second question is 26.90%.

3 0

3 years ago

Which describes the molecule below?

Ira Lisetskai [31]

C. A lipid with three unsaturated fatty acids.
(got the question correct on a test)

4 0

2 years ago

Balance using the redox method:

sweet [91]

Choose all options that apply. Which of the following are equal to 20%? | a) .25 b) 1/5 Oc) 1/10 d) .20

6 0

3 years ago

Experiment ideas I could win a lot of money from school

AleksAgata [21]

I am not all understood but for the school to earn money you can:

make

--a raffle

--lotto

-- yard sale

-- class photo

-- origami for sale or something

-- buffet or food sale (example all Friday ice cream sale, 2 livre ice cream)

6 0

3 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

3 years ago