If the sequence on the right hand side of the DNA molecule was TAGGCTA, the complementary side would have the sequence of what?

A flashbulb of volume 2.70 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 30.0 °C. How many grams of O2(g) doe

Sholpan [36]

P = 2.30 atm

Volume in liter = 2.70 mL / 1000 => 0.0027 L

Temperature in K = 30.0 + 273 => 303 K

R = 0.082 atm

molar mass O2 = 31.9988 g/mol

number of moles O2 :

P * V = n * R* T

2.30 * 0.0027 = n * 0.082 * 303

0.00621 = n * 24.846

n = 0.00621 / 24.846

n = 0.0002499 moles of O2

Mass of O2:

n = m / mm

0.0002499 = m / 31.9988

m = 0.0002499 * 31.9988

m = 0.008 g

3 0

3 years ago

How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?

tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄ → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CO₂: 1 mole
H₄: 4 moles
CH₄: 1 mole
H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

$moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}$

<u><em>moles of CH₄= 340.4 moles</em></u>

Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0

1 year ago

Please help, I really don’t understand this!!!

kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen = 32 / 30

Mass of Oxygen = 1.06 grams

5 0

3 years ago

A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.

sleet_krkn [62]

Explanation:

The given data is as follows.

Pressure (P) = 760 torr = 1 atm

Volume (V) = $720 cm^{3}$ = 0.720 L

Temperature (T) = $25^{o}C$ = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

PV = nRT

Total atoms present (n) = $\frac{PV}{RT}$

= $1 \times \frac{0.720 L}{0.0821 \times 298}$

= 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

x + y = 0.0294

Also, 40x + 131y = 2.966

x = 0.0097 mol

y = (0.0294 - 0.0097)

= 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = $\frac{y}{(x+y)}$

= $\frac{0.0197}{0.0294}$

= 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0

2 years ago

When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio

8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

$V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL$

4 0

3 years ago