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3 years ago

HALLP ME STUCKKKKK

Chemistry

2 answers:

likoan [24]3 years ago

7 0

Answer:

The average height of the sunflower sprouts at the end of week 3 is 12.0

cm.

The average height of the birch sprouts at the end of week 3 is 7.2

cm.

Explanation:

they showed it on edgu

astra-53 [7]3 years ago

4 0

Answer: Sunflower height-12.0 centimeters

Birch Plant-7.2 centimeters

Explanation:

Its how much they grew.

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2. Consider the reaction 2 Cg H18 (4) +250â (9) âº 16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1

Sladkaya [172]

Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

Solution for part (a) : Given,

Moles of $C_8H_{18}$ = 16.15 moles

First we have to calculate the moles of $H_2O$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react to give 18 moles of $H_2O$

So, 16.15 moles of $C_8H_{18}$ react to give $\frac{16.15}{2}\times 18=145.35$ moles of $H_2O$

The moles of water produced are 145.35 moles.

Solution for part (b) : Given,

Mass of $C_8H_{18}$ = 878 g

Molar mass of $C_8H_{18}$ = 114 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_8H_{18}$ .

$\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react with 25 moles of $O_2$

So, 7.702 moles of $C_8H_{18}$ react with $\frac{7.702}{2}\times 25=96.275$ moles of $O_2$

Now we have to calculate the mass of $O_2$ .

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g$

The mass of oxygen needed are 3080.8 grams.

6 0

3 years ago

Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles

Mrrafil [7]

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃ + O₂ ⟶ N₂ + H₂O

The balanced reaction:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

6 0

3 years ago

A sample of Neon is in a sealed container held under isothermic conditions. The initial pressure and volume are 2.7 atm and 4.5

Marina CMI [18]

Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

1.Use Boyle's Law( $P_{1} V_{1}= P_{2} V_{2}$ ). Re-arrange to solve for $V_{2}$ for the final volume.

2. Plug in values. $V_{2} =\frac{(2.7 atm)(4.5 mL)}{(1.7 atm)} = 7.14 mL$

3 0

2 years ago

What is the mass of and object with a volume of 100 cm3 and a density of 10 g/cm3?

nadezda [96]

Answer:

1000 g

Explanation:

d = m/v

We are given d: 10g/cm3

and v: 100cm3

Plug them into the equation to get 10 = m/100

Then, cross multiply 10x100 to get mass which is: 1000g

4 0

2 years ago

A compound decomposes by a first-order process. if 25.0% of the compound decomposes in 60.0 minutes, the half-life of the compou

Tom [10]

Answer:d. 145 minutes

Half-life is the time needed for a radioactive to decay half of its weight. The formula to find the half-life would be:

Nt= N0 (1/2)^ t/h

Nt= the final mass
N0= the initial mass
t= time passed
h= half-life

If 25.0% of the compound decomposes that means the final mass would be 75% of initial mass. Then the half-live for the compound would be:
Nt= N0 (1/2)^ t/h
75%= 100% * (1/2)^ (60min/h)
3/4= 1/2^(60min/h)
log2 3/4 = log2 1/2^(60min/h)
0.41503749928 = -60min/h
h= -60 min / 0.41503749928= 144.6min

3 0

3 years ago