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4 years ago

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Gaseous methane ch4 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 14.0

g of methane is mixed with 86. g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction.

1 answer:

never [62]4 years ago

6 0

<span>2 C2H6 + 7 O2 = 4 CO2 + 6 H2O
I know I got this wrong</span>

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Need help with the question

Sloan [31]

Answer:

for a i think it is choice 2

for b i think it is the first choice

Explanation:

3 0

3 years ago

What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly

Andrej [43]

Answer:

$3.65~M$

Explanation:

We have to remember the <u>molarity equation</u>:

$M=\frac{mol}{L}$

So, we have to calculate "mol" and "L". The total volume is 100 mL. So, we can do the <u>conversion</u>:

$100~mL\frac{1~L}{1000~mL}=~0.1~L$

Now we can calculate the moles. For this we have to calculate the <u>molar mass</u>:

O: 16 g/mol

H: 1 g/mol

C: 12 g/mol

$(16*1)+(1*4)+(12*1)=32~g/mol$

With the molar mass value we can <u>calculate the number of moles</u>:

$1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH$

Finally, we can <u>calculate the molarity</u>:

$M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M$

I hope it helps!

7 0

3 years ago

During which stage of the water cycle could water enter the atmosphere as a gas? A. transpiration B. precipitation C. accumulati

Fantom [35]

Answer: Transpiration---A

Explanation: Transpiration is the process in the water cycle whereby plant loose(excess) water by evaporation through the stomata of their leaves since not all water absorbed by the root is actually used for growth in plants.In order to allow the intake of carbon-dioxide, water must exit the leaves through transpiration which then provides the plant with cooling, rigidity and maintaining the overall water balance of the plant.

8 0

3 years ago

A 135 g sample of a metal requires 2.50 kJ to change its temperature from 19.5°C to 95.5°C to 100.0°C?

just olya [345]

Answer:

0.23J/g°C

Explanation:

Given parameters:

Mass of sample = 135g

Amount of heat = 2.5kJ

Initial temperature = 19.5°c

Final temperature = 100°C

Unknown:

Specific heat capacity of the metal = ?

Solution:

The specific heat capacity of a substance is the amount of heat required to the raise the temperature of 1g of the substance by 1°C.

H = m C (T₂ - T₁ )

H is the amount of heat

m is the mass

C is the specific heat capacity

T₂ is the final temperature

T₁ is the initial temperature;

2.5 x 10³ = 135 x C x (100 - 19.5)

2500 = 10867.5C

C = $\frac{2500}{10867.5}$ = 0.23J/g°C

8 0

3 years ago

Nitrate salts (NO3), when heated, can produce nitrites (NO2) plus oxygen (O2). A sample of potassium nitrate is heated, and the

Lostsunrise [7]

Answer: a) 0.070 moles of oxygen were produced.

b) New pressure due to the oxygen gas is 2.4 atm

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 2.7 atm

V = Volume of gas = 700 ml = 0.7 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = 329 K

$n=\frac{PV}{RT}$

$n=\frac{2.7atm\times 0.7L}{0.0821 L atm/K mol\times 329K}=0.070moles$

Thus 0.070 moles of oxygen were produced.

When the 700 mL flask cools to a temperature of 293K.

$PV=nRT$

$P=\frac{nRT}{V}$

$P=\frac{0.070\times 0.0821\times 293}{0.7}$

$P=2.4atm$

The new pressure due to the oxygen gas is 2.4 atm

7 0

4 years ago