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3 years ago

A 10.0 mL syringe contains 0.10 g of nitrogen gas at 0.0 degrees Celsius. What is the pressure inside the syringe?

1 answer:

4 0

Since the sample inside the syringe is a gas, we assume that it is an ideal gas that is we use the expression PV = nRT. We can easily calculate the pressure as follows:

n = m /MM = 0.10 / 28 = 0.0036 mol

P = nRT / V = 0.0036(0.08206)(273.15) / 0.010
P = 8 atm

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One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2.

mart [117]

Answer:

You could expect 3.48 grams of C₂H₄N₂

Explanation:

You must start by stating the chemical equation for the reaction of ammonia, carbon dioxide, and methane to produce aminoaceto nitrile.

1. Word equation:

Ammonia + Carbon dioxide + Methane → Aminoacetonitrile + Water

2. Balanced chemical equation:

$8NH_3+5CO_2+3CH_4\rightarrow 4C_2H_4N_2 +10H_2O$

3. Convert the mass of each reactant into number of moles:

Formula:

Number of moles = mass in grams/molar mass

2.11g NH₃

Number of moles = 2.11g / 17.03g/mol = 0.124 mol NH₃

14.9g CO₂

Number of moles = 14.9g/44.01g/mol = 0.339 mol CO₂

1.75g CH₄

Number of moles = 1.75g/16.04g/mol = 0.109 mol CH₄

4. Theoretical mol ratio

From the balanced chemical equation, using the coefficientes:

$8molNH_3:5molCO_2:3molCH_4:4molC_2H_4N_2:10molH_2O$

5. Limiting reagent

The available amounts of the reactants are:

0.124 mol NH₃

0.339 mol CO₂

0.109 mol CH₄

Fom the theoretical mole ration, to react with 0.124 mol of NH₃ you would need:

0.124molNH₃ × (5molCO₂/8molNH₃) = 0.0775 mol CO₂

Since there are 0.339 moles available, this is in excess.

0.124molNH₃ × (3molCH₄/8molNH₃) = 0.0465mol CO₂

Since there are 0.109 moles available, this is in excess too.

Hence, the limiting reagent is NH₃.

6. Yield

Use the theoretical ratio:

0.124molNH₃ × (4molC₂H₄N₂ / 8molNH₃) = 0.0620 mol C₂H₄N₂

Convert to grams:

Mass = number of moles × molar mass

0..0620 mol × 56.068g/mol = 3.48 g of C₂H₄N₂ ← answer

4 0

4 years ago

Molar mass (NH4)2SO4

Hoochie [10]

Answer:

132.15

Explanation:

Molar mass N = 14.00

Molar mass H = 1.01

Molar mass H4 = 1.01 x 4 = 4.04

Molar mass NH4 = 14.00 + 4.04 = 18.04

Molar mass (NH4)2 = 18.04 x 2 = 36.08

Molar mass S = 32.07

Molar mass O = 16.00

Molar mass O4 = 16.00 x 4 = 64.00

Molar mass SO4 = 32.07 + 64.00 = 96.07

Molar mass (NH4)2SO4 = 36.08 + 96.07 = 132.14

6 0

3 years ago

100 PIONTSSSSS HELP ASAP

valina [46]

Left Panel

Short answer A

Solution

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

Step One

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

Step Two

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

Step Three

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

Step Four

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

Step Five

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P = 2 * 31 = 62

8 O = 8 * 16 =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

Step Six

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

Step One

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C

6 0

3 years ago

How can chemical changes be detected

saul85 [17]

Rarely they can't with just sight. Certain tests or experiments should take place

7 0

4 years ago

How does the plate boundary along the west coast of the United States change over time?

Tanzania [10]

The North American plate is moving towards the west-southwest at about 2.3 centimeters every year mediated by the Mid-Atlantic Ridge, the spreading center, which gave rise to the Atlantic Ocean. The small Juan De Fuca plate, moving east-northeast at 4 centimeters every year, was once a component of much greater oceanic plates known as the Farallon plate.

The Farallon plate used to comprise what is now the Cocos plate of Mexico and Central America, and the Juan de Fuca plate in the region from N. Vancouver Island to the Cape Mendicino California, and a big sea floor tract in between. However, the middle portion of the Old Farallon plate disappeared underneath North America, it was subducted underneath California leaving the San Andreas fault system behind as the contact between the Pacific plates and North America.

The Juan De Fuca plate is still actively subducting underneath North America. Its movement is not smooth, however, rather sticky. The buildup of strain takes place until the fault dissociates and a few meters of Juan De Fuca get slid underneath North America in a big earthquake.

7 0

3 years ago

Read 2 more answers