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galina1969 [7]
3 years ago
13

OP

Chemistry
2 answers:
Yuri [45]3 years ago
8 0

the answer is d hope this helps

bogdanovich [222]3 years ago
4 0

Answer:

The answer is d

Explanation:

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What is the particle that is labeled with a question mark (?) in the diagram?
Nat2105 [25]

Answer is: quark.

Quark is a type of elementary particle and a fundamental constituent of matter.

Quarks form composite hadrons (protons and neutrons). Protons and neutrons are in the nucleus of an atom.

Hadrons include baryons (protons and neutrons) and mesons.

There are six types of quarks: up, down, strange, charm, bottom, and top.

5 0
4 years ago
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How many grams are in 1 mole of Pb3(PO4)4?
kvasek [131]

Answer:

In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams

Explanation:

This compound is the lead (IV) phosphate.

Grams that occupy 1 mole, means the molar mass of the compound

Pb = 207.2  .3 = 621.6 g/m

P = 30.97 .4 = 123.88 g/m

O = (16 .  4) . 4 = 256 g/m

621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m

6 0
3 years ago
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What is the oxidation state of Hg in Hg2Cl2?​
Anna [14]

Answer:

+1

Explanation:

Electrochemistry. In oxidation–reduction (redox) reactions, electrons are transferred from one A redox reaction is balanced when the number of electrons lost by the reductant Hg(l)∣Hg2Cl2(s)∣Cl−(aq) ∥ Cd2+(aq)∣Cd(s).

As is evident from the Stock number, mercury has an oxidation state of +1. This makes sense, as chlorine usually has an oxidation state of -1.

5 0
3 years ago
Someone help please.
mojhsa [17]

After 3 half life periods you would have 5 grams of krypton left because half of 40 is 20 half of 20 is 10 and half of 10 is 5

6 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
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