Chemistry College

How much excess reactant remains from the reaction of 8.3g of sodium and 4.5g of chlorine?
a. 8.2g Na
b. 5.4g Na
c. 5.4g Cl2
d. 8.2g Cl2

Answers

Answer 1
Answer: I wanna say A but I’m not 100% sure
But I hope this helps at least a little :)
Answer 2

Answer:

5.4 g Na

Explanation:

Calculate the mass of sodium chloride from both reactants. The lesser value is the amount is the theoretical yield. Subtract the mass of the calculated mass of sodium chloride from the mass of the reactants.

Related Questions

Which of the following statements is true?

Question 10 options:

a chromosome is larger than a cell


a DNA strand is made of many chromosomes


a gene contains may chromosomes


a chromosome contains many genes

Answers

Answer:

a chromosome contains many genes

Explanation:

Polonium-214 went through a series of radioactive decays to produce a stable isotope of bismuth-210 choose the correct decay series that would produce bismuth-210
A: Polonium-214 undergoes 2 beta decay and 1 gamma emission
B: Polonium-214 undergoes 1 alpha decay and 2 gamma emission
C: Polonium-214 undergoes 2 alpha decay and 1 beta decay
D: Polonium-214 undergoes 1 alpha decay 1 beta decay and 1 gamma emission.

Answers

The correct decay series :

₈₄Po²¹⁴⇒₈₃Bi²¹⁰+₂He⁴+ ₋₁e⁰+ ₀γ⁰

Further explanation

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,  

alpha α particles ₂He⁴ beta β ₋₁e⁰ particles gamma particles ₀γ⁰ positron particles ₁e⁰ neutron ₀n¹

Decay's reaction :

₈₄Po²¹⁴⇒₈₃Bi²¹⁰+x+y+z

From this decay reaction can be concluded

mass number decreased by 4

atomic number decreased by 1

Then the most suitable are:

Polonium-214 undergoes 1 alpha decay 1 beta decay and 1 gamma emission.

₈₄Po²¹⁴⇒₈₃Bi²¹⁰+₂He⁴+ ₋₁e⁰+ ₀γ⁰

So that the mass number and atomic number are the same both before and after decay

Pro
∆G0=-RT in k Known as​

Answers

Answer:

Gibbs free energy equation

Explanation:

A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution?
A) 0.126g
B) 3.66g
C) 4.03g
D) 3.84g
E) 280 g

Answers

Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume =  100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

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