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Mila [183]
2 years ago
10

Neutrons have neither charge nor mass. TRUE FALSE

Chemistry
1 answer:
iren2701 [21]2 years ago
7 0

Answer:

FALSE

nuetrons do indeed have no charge, however they are nearly 2000 more times massive than electrons.

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How much water, in grams, can be made from 1.03 x <img src="https://tex.z-dn.net/?f=10%5E%7B24%7D" id="TexFormula1" title="10^{2
krok68 [10]

Answer:

30.8 g of water are produced

Explanation:

First of all we need the equation for the production of water:

2H₂ + O₂ → 2H₂O

2 moles of hydrogen react with 1 mol of oxygen in order to produce 2 moles of water.

As we assume, the oxygen in excess, we determine the moles of H₂.

1.03ₓ10²⁴ molecules . 1 mol/ 6.02ₓ10²³ molecules = 1.71 moles

Ratio is 2:2, so 1.71 moles will produce 1.71 moles of water

Let's convert the moles to mass: 1.71 mol . 18g / 1mol = 30.8 g of water are produced

5 0
3 years ago
How many particles are in 697.008 grams of Potassium Sulfate?
g100num [7]

Answer:

174.2592

Explanation:

Trust me.

8 0
3 years ago
Do plasmids have an importance beyond the practice of genetic engineering?
kozerog [31]
Answer: Yes

Explanation: Plasmids offer a number of unique characteristics that make genetic engineering much more efficient. Plasmids are a type of non-chromosomal DNA. Integrating DNA into a bacterial or other chromosome is far more complex than simply putting DNA into a cell; plasmids make it easier to transport DNA into a cell by eliminating this step.

4 0
2 years ago
What's the Balanced equation for C5H11OH+O2 ​
Galina-37 [17]

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8 0
3 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
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