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3 years ago

Which of the following is not produced in

Chemistry

2 answers:

aalyn [17]3 years ago

5 0

The one that is not produced in fission reactions is (1) Greenhouse gases

( Whisper)BTW i searched it up :)-

Anastaziya [24]3 years ago

3 0

1. greenhouse gases. these are not formed because nuclear fission involves the modification of atoms. It does not give rise to other molecules.

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dsp73

Long wavelengths and low frequencies

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2 years ago

What is the most basic level of biological organization?

Brums [2.3K]

Answer:

molecule

Explanation:

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2 years ago

Read 2 more answers

What does index fossil mean?

marysya [2.9K]

An index fossil is a fossil that is useful for dating and correlating the strata in which it is found.

Or it is a fossil that helps you find the date where it has been found. I rephrased that.

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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

A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027

irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

$C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\$

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

x _________ 25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

7 0

3 years ago