1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
laiz [17]
3 years ago
8

What are you and what do you want with my attendance

Chemistry
1 answer:
Lelu [443]3 years ago
8 0

Answer:

Such a waste of points but lol

You might be interested in
While finding the number of molecules of oxygen molecules present in 3.65 moles of Na2SO4 the conversion factor used would be
zvonat [6]

Answer

Avogadro's number: One mole of any substance contains 6.022×10²³ molecules

Explanation

While finding the number of moles of oxygen molecules present in 3.65 moles of Na2SO4 the conversion factor used would be Avodagro's number, which is

One mole of any substance contains 6.022×10²³ molecules.

8 0
1 year ago
CARBOHYDRATES are found primarily in plant-based foods and are a major source of energy in the body (4 kcals per gram). If you c
SVETLANKA909090 [29]

Answer:

The 200 grams of carbohydrate would this convert to 800 kcals in a day.

Explanation:

Carbohydrates are the source of energy. The metabolism of 1 gram of carbohydrates gives 4 kcals of energy.

If 200 grams of carbohydrate is consumed in a day, then the energy provided by it is:-

Energy\ provided\ by\ the\ carbohydrate = 200\times 4\ kcals=800\ kcals

<u>The 200 grams of carbohydrate would this convert to 800 kcals in a day.</u>

4 0
2 years ago
Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (
Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0
3 years ago
Perform the following operation
lukranit [14]

Answer:

(5.4 x 10³) x (1.2 x 10⁷) = 6.48 x 10¹⁰

With correct significant figures, the answer would be 6.5 x 10¹⁰.

4 0
2 years ago
The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the rea
timofeeve [1]

Answer:

K_{p} for the reaction is 18.05

Explanation:

Equilibrium constant in terms of partial pressure (K_{p}) for this reaction can be written as-

                K_{p}=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}

where P_{NO_{2}} and P_{N_{2}O_{4}} are equilibrium partial pressure of NO_{2} and N_{2}O_{4} respectively

Hence K_{p}=\frac{(0.095)^{2}}{(0.0005)} = 18.05

So, K_{p} for the reaction is 18.05

3 0
3 years ago
Other questions:
  • What is the mol of 1 5 x 10^26
    12·1 answer
  • Give the characteristics of a strong acid.
    13·1 answer
  • a student is studying the ways different elements are similar to one another diagrams of atoms from four different elements are
    7·1 answer
  • What is the least dense layer of the earth
    13·2 answers
  • A gas is initially at a pressure of 0.43 atm, and a volume of 11.7 liters. Then the pressure is raised to 3.61 atm and the volum
    12·2 answers
  • Isotopes of the same element have a different number of ______​
    14·1 answer
  • A student added white crystals to a clear liquid. The crystals dissolved in the liquid. After a few seconds, the liquid became q
    11·1 answer
  • What is a cation and where can you find it on the periodic table?
    7·1 answer
  • Analyze How would tides be affected if the<br> Moon was farther away from Earth?
    15·2 answers
  • How much would the boiling point of water increase if mol of NaCl added 1 kg of water (Kb= 0.51 degrees Celsius /(mol/kg) for wa
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!