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3 years ago

Can anyone please help ME with Question 1

Chemistry

2 answers:

Flauer [41]3 years ago

4 0

Enzymes are important because they speed up chemical reactions that take place in within cells.

Hope this helped! :D

Vsevolod [243]3 years ago

3 0

Enyzmes speed up all chemical reactions

(Just took an AP Bio test with this question on it)

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Which of the following pairs of elements is most likely to form an ionic bond? A. nitrogen and oxygen B. aluminum and nitrogen C

krok68 [10]

Answer:

Explanation:

When nitrogen and aluminum form an ionic bond, the formula of the ionic compound is AlN. Aluminum forms a cation Al⁺³ and nitrogen forms an anion N⁻³. When each cation of Al⁺³ combines with an anion of N⁻³, AlN or aluminum nitride is formed

\Hope i could help xxxxxxxxxxxxxxxxxxx

6 0

3 years ago

An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

melisa1 [442]

570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students

7 0

3 years ago

Read 2 more answers

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

What is the partial pressure of carbon dioxide in a container that contains 3.63 mol of oxygen, 1.49 mol of nitrogen, and 4.49 m

lana66690 [7]

Answer:

Partial pressure of CO₂ is 406.9 mmHg

Explanation:

To solve the question we should apply the concept of the mole fraction.

Mole fraction = Moles of gas / Total moles

We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)

Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles

To determiine the partial pressure of CO₂ we apply

Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P

Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure

We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg

6 0

3 years ago

Read 2 more answers

Calculate the number of atoms in 35 g of carbon. Calculate the number of atoms in 75 g of bromine.

nata0808 [166]

Answer:

Explanation:

To calculate the number of atoms in a formula, the weight of a sample, its atomic mass from the periodic table and a constant known as Avogadro’s number are needed.

Step 1: Find the Molar Mass of the Formula

Find a periodic table of elements to find the molar mass of your sample. If your sample is made of one element, like copper, locate the atomic mass on the periodic table. Atomic mass is usually listed below the symbol for that element.

The atomic mass of carbon and bromine is 63.55 atomic mass units. This formula mass is numerically equal to the molar mass in grams/mole, and this means copper is 63.55 grams/mole.

Whether you use an individual element like copper or a molecule, the procedure for finding the atoms in a formula remains the same.

Find the molar mass of carbon and bromine on the periodic table: 63.55 grams/mole.

Understanding the Mole: The mole (often abbreviated as mol) listed above is a unit of measurement. If you sold eggs, you would talk about them in the dozens, not one by one.

A mole is a certain amount, too. If chemists want to speak about incredibly small atoms and molecules, an amount far greater than a dozen is needed. A mole is Avogadro’s number of items: 6.022 × 1023.

1 mole of C atoms = 6.022 × 1023 C atoms

1 mole of Br atoms = 6.022 × 1023 S atoms

1 mole of CBrmolecules = 6.022 × 1023 CBr molecules

1 mole of pennies = 6.022 × 1023 pennies

To give an idea of how large this number is, 1 mole of pennies would be enough money to pay all the expenses of each country on earth for about the next billion years.

Step 2: Find the Number of Moles

The example is 35 grams of C and . Change that into moles using the molar mass you found in Step 1. Chemists use ratios for this calculation.

Start with what you know and add in the molar mass ratio, so the units will cancel:

35g of C× 1 mol C/ 12 g C = 2.92mol of C

Step 3: Convert Moles to Atoms Using Avogadro’s Number

Once the amount of moles is known, the number of atoms in the formula may be calculated using Avogadro’s number. Again, use the ratio format.

Notice the number of moles is used from Step 2 to start the calculation from moles to atoms:

2.92 mol of C× 6.022 x 1023 atoms / 1 mol of C = 3.13 x 1023 molecules

To answer your example question, there are 3.13 × 1023 atoms in 32.80 grams of carbon.

Steps 2 and 3 can be combined. Set it up like the following:

32.80 g of C × 1 mol Cu / 159.17 g C × 6.022 x 1023 atoms / 1 mol of C = 3.13 x 1023 atoms in 32.80 grams of carbon

Several online sites have a number of atoms calculator. One is the Omni Calculator and is listed in the Resources section, but you’ll still need to know how to calculate molar mass (Step 1).

Historical Note: Why Is It Called Avogadro's Number?

Avogadro’s number is named after Amedeo Avogadro (1776-1856), an Italian scientist that hypothesized that equal volumes of gases at the same temperature and pressure will have the same number of particles.

Avogadro did not propose the constant, 6.022 ×1023, but because of his contributions to science, the constant was named after him. Incidentally, Avogadro first introduced his gas theory in 1811, and it was ignored for 50 years.

5 0

3 years ago