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Alexus [3.1K]
3 years ago
7

Please help me with this

Chemistry
1 answer:
Elina [12.6K]3 years ago
4 0

Answer:

Speed and acceleration

Explanation:

i would say this bc the formula is related to acceleration if you looking at all the possible exceptions im not a expert in this range but im just giving you what i would pick so if you get it wrong im sorry lol.

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Dr. Terror has developed a new alloy called Ultranomium. He is testing a bar that is 1.25 m1.25 m long and has a mass of 382 g.3
MrRissso [65]

Answer:0.8742j/g°C

Explanation: SOLUTION

GIVEN

length of bar=1.25m

mass 382g

temperature= 20°C to 288°C

Q=89300J

Specific Heat Capacity  will be calculated using

Q=mC∆T

where

C = specific heat capacity

Q = heat

m = mass

Δ T = change in temperature

C=Q/ m∆T

=89300/382X(288-20.6)

=0.8742j/g°C

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How do you find the number of valence electrons
Ganezh [65]

Answer:LULULLUL

Explanation:

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3 years ago
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Predict the products and write the balanced chemical equation for this reaction. Include the physical state of the product(s).
sukhopar [10]

Answer:

Al(C2H3O2)3 (aq) + 3AgNO3 (aq) ------>Al(NO3)3(aq) + 3Ag(C2H3O2)(s)

Explanation:

A chemical reaction refers to an interaction between two or more chemical species that leads to the formation of other new chemical species.

There are many types of chemical reactions. The type shown above is called a double replacement reaction. In this type of reaction, ions exchange partners in the product.

A chemical reaction is said to be balanced when the number of atoms of each element on the left hand and right hand sides of the reaction equation are exactly the same.

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larisa [96]

<span>fragmentation that is the answer</span>
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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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