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3 years ago

11

In a chemical reaction, the __________ is consumed completely during the reaction. A) stoichiometry B) excess reagent C) ideal c

ompound D) limiting reactant

1 answer:

Anuta_ua [19.1K]3 years ago

7 0

I believe it is D. because stoichiometry allows people to determine how much reactant is consumed in a chemical reaction.

You might be interested in

How many moles of oxygen are required to produce 37.15 g CO2? 37.15 g CO2 = mol O2

NNADVOKAT [17]

Answer:

0.84 moles of oxygen are required.

Explanation:

Given data:

Mass of CO₂ produced = 37.15 g

Number of moles of oxygen = ?

Solution:

Chemical equation:

C + O₂ → CO₂

Number of moles of CO₂:

Number of moles = mass/molar mass

Number of moles = 37.15 g/ 44 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of oxygen and carbon dioxide.

CO₂ : O₂

1 : 1

0.84 : 0.84

0.84 moles of oxygen are required.

6 0

3 years ago

Read 2 more answers

Iron has density 7.87 g/cm^3. if 52.4 g of iron is added to 75.0 mL of water in a graduated cylinder, to what volume reading wil

vlada-n [284]

Answer:

6.66 mL

Explanation:

The increase in the volume is due to the addition of the iron whose volume can be calculated as:

Using,

Density = Mass / Volume

Given that:

Density of Iron = 7.87 g/cm³

Mass of iron = 52.4 g

Thus, volume is:

Volume = Mass / Density = 52.4 / 7.87 cm³ = 6.66 cm³

Also, 1 cm³ = 1 mL

<u>The rise in the volume = 6.66 mL</u>

8 0

3 years ago

The mass percent of carbon in a typical human is 18%, and the mass percent of 14C in natural carbon is 1.6 × 10-10%. Assuming a

Tema [17]

Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

$\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g$

Carbon-14 is $1.6\times 10^{-10}\ \%$ of the carbon in the body. So,

$\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g$

Also,

14 g of Carbon-14 contains $6.023\times 10^{23}$ atoms of carbon-14

So,

$2.48\times 10^{-8}\ g$ of Carbon-14 contains $\frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8}$ atoms of carbon-14

Atoms of carbon-14 = $1.07\times 10^{15}$

Given that:

Half life = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.00012 years⁻¹

Also, 1 year = $3.154\times 10^7$ s

So, The rate constant, k = $\frac{0.00012}{3.154\times 10^7}$ s⁻¹ = $3.8\times 10^{-12}\ s^{-1}$

Thus, decay events per second = $K\times atoms decayed$ = $3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s$ = 4066 decay/s

4 0

3 years ago

Give examples of temporaty physical change

garri49 [273]

ice because it can melt back to water

3 0

3 years ago

zhenek [66]

Answer:

Air pollution

Explanation:

Carbon(IV)oxide is sometimes harmful to man but good for plant in other to manufacture their food

6 0

3 years ago