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IgorC [24]
3 years ago
14

What is the resistance of a blub when the voltage across is 6 V and the current is 0.2A?

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

30 Ohms (Ω)

Explanation:

Resistance = Voltage/Current

R = 6/0.2

R = 30 Ohms (Ω)

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Calculate the percent by mass of a solution of Ca(NO3)2 that has 22.63 g dissolved in 896.92 g of water.
never [62]

Hey there!

To calculate the percent by mass of the Ca(NO₃)₂ we need to find the total mass first by adding.

896.92 + 22.63 = 919.55

In total, the solution is 919.55 grams.

To find the percent of Ca(NO₃)₂ in the solution, divide the mass of Ca(NO₃)₂ by the total mass and multiply by 100.

22.63 ÷ 919.55 = 0.0246

0.0246 x 100 = 2.46

Ca(NO₃)₂ makes up 2.46% of the solution.

Hope this helps!

6 0
3 years ago
Given the following unbalanced equation:
antiseptic1488 [7]
<h3>Answer:</h3>

11.84 mol CoF₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[Given] 11.84 moles CoCl₂

[Solve] moles CoF₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol CoCl₂ → 1 mol CoF₂

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                       \displaystyle 11.84 \ mol \ CoCl_2(\frac{1 \ mol \ CoF_2}{1 \ mol \ CoCl_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 11.84 \ mol \ CoF_2
7 0
3 years ago
8. What group numbers are the transition elements in?
Anna007 [38]
The 38 elements in groups 3 through 12 of the periodic table are called "transition metals". As with all metals, the transition elements are both ductile and malleable, and conduct electricity and heat. The interesting thing about transition metals is that their valence electrons, or the electrons they use to combine with other elements, are present in more than one shell. This is the reason why they often exhibit several common oxidation states. There are three noteworthy elements in the transition metals family. These elements are iron, cobalt, and nickel, and they are the only elements known to produce a magnetic field I HOPE THIS HELP:)
3 0
3 years ago
Use MO diagrams to place B2+, B2, and B2- in order of (a) decreasing bond energy; (b) decreasing bond length.
shepuryov [24]

We use the MO diagram for a homonuclear diatomic species (since C and N are neighbours, we treat them as the "same").

The first two electrons contribute to bonding. The next two are anti-bonding.

The next six contribute to bonding, and the following six are anti-bonding.

So, if we start with CN+, which has 4+5-1 (8) valence electrons, we note that the first two electrons contribute to bonding, while the next two cancel this out; the next four contribute to bonding, so the bond order is 4/2 = 2.

If we add one more electron to get CN, there are now 5 bonding electrons, giving bond order 5/2=2.5.

Adding one more to give CN- would give the bond order 6/2 = 3. (If we added more electrons, each one would lower the bond order.)

Given a series of molecules with identical skeletal structures, the one with the highest bond order has the highest bond energy:

CN+ < CN < CN-

Lewis structures will verify that CN- has a triple bond, but they do not work particularly well for CN+ and CN.

learn more about bond orders at

brainly.com/question/9713842

#SPJ1

3 0
1 year ago
A soft silvery metal has two naturally occurring isotopes: mass 84.9118, accounting for 72.15% and mass 86.9092, accounting for
zloy xaker [14]

Answer: The atomic weight of the metal would be 85.47.

Explanation:

Mass of isotope 1 of metal = 84.9118

% abundance of isotope 1 of metal = 72.15% = \frac{72.15}{100}

Mass of isotope 2 of metal= 86.9092

% abundance of isotope 2 of metal = 27.85% = \frac{27.85}{100}

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(84.9118)\times \frac{72.15}{100})+(86.9092)\times \frac{27.85}{100}]]

A=85.47

Therefore, the atomic weight of the metal would be 85.47.

6 0
3 years ago
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