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lys-0071 [83]
1 year ago
11

How many significant figures are in this number? 43.55

Chemistry
1 answer:
leva [86]1 year ago
7 0

Answer:

4 significant figures

Explanation:

Significant figures are the units/digits within a number that make the number more accurate and precise.

All digits (except for 0) are always significant. Therefore, all the digits in 43.55 are significant. Since there are 4 digits in the given number, there are 4 significant figures.

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Energy from a reaction was determined to be 3870 J using a water calorimeter, when burning 9.00 grams of H­2. What is the total
Paraphin [41]

Answer:

860 J / mol

Explanation:

Enthalpy = Energy / no. of moles

no. of moles = mass / molar mass

Take the atomic mass of H = 1.0,

molar mass of H2 = 1.0 x 2

= 2.0

no. of moles of H2 = 9.00/2

= 4.5 mol

Hence,

Enthalpy = 3870 / 4.5

=860 J / mol

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Coastal waves dissolving limestone sediments
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Answer:

Physical Change

Explanation:

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umka2103 [35]

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3 years ago
Read 2 more answers
TRY H Vele
Strike441 [17]
<h3>Answer:</h3>

Al- [Ne] 3s²3p¹

As- [Ar] 4s²3d¹⁰ 4p³

Explanation:

  • Electron configuration of an element shows the arrangement of electrons in the energy levels or orbitals in the atom.
  • Noble-gas configuration involves use of noble gases to write the configuration of other elements.
  • This is done by identifying the atomic number of the element and then identifying the noble gas that comes before that particular element on the periodic table.
  • For example:
  1. Aluminium: The atomic number of Al is 13. The noble gas before Aluminium is Neon which has 10 electrons. Therefore the remaining 3 electrons fills up the 3s and 3p sub orbitals.
  • Thus, the noble-gas configuration of Al is [Ne] 3s²3p¹

     2.  Arsenic, Atomic number is 33

  • Noble gas before Arsenic is Ar,. Argon has 17 electrons, then the remaining electrons fills up the 4s, 3d and 4p sub-orbitals.
  • Thus, the noble-gas configuration of As is [Ar] 4s²3d¹⁰ 4p³
6 0
2 years ago
(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of
Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

P x V = k

Therefore, for this problem the step by step explanation is:

P_{1} xV_{1} = P_{2} xV_{2}

Clearing P2 and replacing

P_{2}= \frac{P_{1} xV_{1}}{V_{2} } = \frac{4atmx3L}{20L} = 0.6atm

7 0
3 years ago
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