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2 years ago

Please help me out i will give you brainlist. 0.500 is wrong

Chemistry

2 answers:

SOVA2 [1]2 years ago

8 0

Answer:

.250

Explanation:

Alik [6]2 years ago

3 0

<h3>Answer: b) 0.250 mol</h3>

============================================

Work Shown:

Using the periodic table, we see that

1 mole of carbon = 12 grams
1 mole of oxygen = 16 grams

These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.

So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.

-------------------

Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.

11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2

In short,

11.0 grams of CO2 = 0.250 mol of CO2

This is approximate.

We don't need to use any of the information in the table.

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a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

2 years ago

1. Some athletes have as little as 3.0% body fat. If such a person has a body mass of 65 kg, how many pounds of body fat does th

77julia77 [94]

1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs

$65 kg \frac{3.0 kg}{100 kg} x \frac{2.20 lb}{1 kg} = 4.29 lbs$

2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams

$3 Liters \frac{1000 mL}{1 L} x \frac{0.94 grams}{1 mL} x \frac{1 kg}{1000 g} x \frac{2.20 lbs}{1 kg} = 6.20 lbs$

3 0

3 years ago

Read 2 more answers

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

The void fraction of the bed is 0.37.

6 0

3 years ago

2a. Write down the names of the new substances formed when each of the following substance reacts with dilute hydrochloric acid.

Naddika [18.5K]

Answer:

i) Dilute hydrochloric acid will react with Ammonia to form ammonia salt.

ii) dilute hydrochloric acid will react with soduim hydroxide to form sodium chloride and water

iii) dilute hydrochloric acid will react with calcuim carbonate to form Calcium chloride, Carbon dioxide and water.

Explanation:

CHEMICAL EQUATIONS :

I) HCL + NH3 = NH4Cl

ii) HCL+ NaOH = NaCl + H2O

iii) HCL + CaCo = CaCl2 + CO2 + H2O

7 0

2 years ago

Read 2 more answers