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3 years ago

Which of these terms refers to matter that could be heterogeneous?

Chemistry

1 answer:

Dafna11 [192]3 years ago

5 0

Answer:

B) mixture

Explanation:

Many homogeneous mixtures are commonly referred to as solutions. A heterogeneous mixture consists of visibly of three phases or states of matter are gas, liquid, and solid.

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I’m making a transition from stare n=1 to state n=2 the hydrogen atom must

MatroZZZ [7]

Answer:

Water, 35 liters. Carbon, 20 kilograms. Ammonia, 4 liters. Lime, 1.5 kilograms. Phosphorous, 800 grams. Salt, 250 grams. Saltpeter, 100 grams. Sulfur, 80 grams. Fluorine, seven-point-five. Iron, five. Silicon, three grams. And trace

amounts of 15 other elements.

the ingredients of the average adult,right down to the last specks of protein in your eyelashes. And even though science has given us the entire physical breakdown, there's never been a successful attempt at bringing a human to life. There's still something missing. Something scientists haven't been able to find in centuries of research. ...and in case you're wondering, all those ingredients can be bought on a child's allowance. humans can be made rather cheap. There's no magic to it.

6 0

3 years ago

A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life

mixas84 [53]

T is amount after time t
Ao is initial amount 
t is time 
HL is half life 

log (At) = log [ Ao x (1/2)^(t/HL) ] 
log (At) = log Ao + log (1/2)^(t/HL) 
log (At) = log Ao + (t/HL) x log (1/2) 

( log At - log Ao) / log (1/2) = t / HL 
log (At/Ao) / log (1/2) = t / HL 

HL = t / [( log (At / Ao)) / log (1/2) ] 

HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] 

HL = 14.4 s / 2 = 7.2 seconds

8 0

3 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aniline:

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

3 years ago

When the molecules of gases are heated, the average kinetic energy of the molecules increases,

serious [3.7K]

True is the answer to this

6 0

3 years ago

How would you prepare 100.0 ml of.400 m CaCl2 from a stock solution of 2.00 M CaCl2?

docker41 [41]

C₀=2 mol/l
c₁=0.400 mol/l
v₁=100.0 ml = 0.1 l

c₁v₁=c₀v₀

v₀=c₁v₁/c₀
v(H₂O)=v₁-v₀

v₀=0.1*0.400/2=0.02 l = 20 ml
v(H₂O)=100 - 20 = 80 ml

It is necessary to mix 20 ml of the feed solution and 80 ml of water.

4 0

3 years ago

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