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ElenaW [278]
3 years ago
9

A basketball weighs approximately 1.35 kg. What is this mass in grams (g)? step by step

Chemistry
1 answer:
choli [55]3 years ago
8 0

Answer:

1350 g

Explanation: just add a 0

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How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5
NeTakaya

Answer:

c

Explanation:

How many moles of gold are equivalent to 1.204 × 1024 atoms?

0.2

0.5

2

5

C) 2 Is the correct answer, I took the test and it was correct.

5 0
3 years ago
Read 2 more answers
What did the scientists have to consider before releasing the dragonflies into the wild? Why?
pentagon [3]
Maybe they had to consider the habitat to make sure the habitat they were releasing the dragonflies into would be appropriate for the dragonflies.
3 0
3 years ago
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(5points) Na2[Fe(OH2)2(OH)I3]absorbs photons with energy of 191kJ/mol. Calculate the wavelength (in nm) for a photon absorbed by
AfilCa [17]

Answer:

626.7nm

Explanation:

The energy of a photon is defined as:

E = hc / λ

<em>Where E is the energy of the photon, h is Planck constant (6.626x10⁻³⁴Js), c is speed of light (3x10⁸m/s) and </em>λ is the wavelength of light

The energy of 1 photon is:

(191000 J / mol) ₓ (1 mole / 6.022x10²³) = 3.1717x10⁻¹⁹ J

Replacing:

3.1717x10⁻¹⁹ J = <em>6.626x10⁻³⁴Jsₓ3x10⁸m/s /  </em>λ

λ = 6.267x10⁻⁷m

as 1nm = 1x10⁻⁹m:

6.267x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =

<h3>626.7nm</h3>
6 0
3 years ago
Calculate how many moles of element Q are in 23.53 g of element Q.
pogonyaev

Answer:

0.579 moles

Explanation:

Moles = 23.53/40.64

= 0.5789 moles

7 0
3 years ago
If you begin with 2.7 g Al and 4.05 g Cl2, what mass of AlCl3 can be produced?
Tresset [83]
<span>atomic weights: Al = 26.98, Cl = 35.45 In this reaction; 2Al = 53.96 and 3Cl2 = 212.7 Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed. Step 2: (a) Ratio of Al:Cl = 2.70/4.05 = 0.6667 since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537. so Cl is limiting (b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced. From Step 1: 212.7g of Cl will produce 266.66g AlCl3 212.7g = 266.66g 4.05g = x x = 5.08g of AlCl3 can be produced (c) Al:Cl = 0.2537 Al:Cl = Al:4.05 = 0.2537 mass of Al used in reaction = 4.05 x 0.2537 = 1.027g Excess reactant = 2.70 - 1.027 = 1.67g King Leo · 9 years ago</span>
8 0
3 years ago
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