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3 years ago

For the reaction

Chemistry

1 answer:

azamat3 years ago

7 0

Answer:

Mass = 5.56 g

Explanation:

Given data:

Mass of Cl₂ = 4.45 g

Mass of NaCl produced = ?

Solution:

Chemical equation:

2Cl₂ + 4NaOH → 3NaCl + NaClO₂ + 2H₂O

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.45 g/ 71 g/mol

Number of moles = 0.063 mol

Now we will compare the moles of Cl₂ with NaCl.

Cl₂ : NaCl

2 : 3

0.063 : 3/2×0.063 =0.095 mol

Mass of NaCl:

Mass = number of moles × molar mass

Mass = 0.095 mol × 58.5 g/mol

Mass = 5.56 g

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Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide tetrahydrate in enough water to make 450 mL

Vladimir [108]

Answer:

The molarity of the CoBr2•4H2O solution is 7.64 × 10-2 M

Explanation:

Cobalt (II) bromide tetrahydrate

• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+

• Bromide - anion from group 7A → -1 charge → symbol: Br-

• Tetrahydrate- tetra- means 4 and hydrate is H2O

The chemical formula of the compound is: CoBr2•4H2O

We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.

We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.

•The molar mass of CoBr2•4H2O is:

CoBr2•4H2O

1 Co x 58.93 g/mol Co = 58.93 g/mol

2 Br x 79.90 g/mol Br = 159.80 g/mol

8 H x 1.008 g/mol H = 8.064 g/mol

4 O x 16.00 g/mol O = 64.00 g/mol

________________________________________

Sum = 290.79 g/ mo

The moles of CoBr2•4H2O is:

= 10.0 g CoBr2•4H2O x $\frac{ 1 mol CoBr_2 . 4H_2O}{290.79 g CoBr_2 . 4H_2O}$

= 0.0344 mol CoBr2•4H

We know that the volume of the solution is 450 mL.

We can now calculate for molarity:

Convert mL to L → 1 mL = 10-3 L

Formula:

Molarity (M)= Mole of solute / Liters of solution

= 0.0344 mol CoBr2•4H / 450 mL x 1 ml / 10^ -3 L

= 0.0764

= 7.64 × 10-2 mol/L

8 0

4 years ago

How many grams of O2 are present in 44.1 L of O2 at STP?

ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ 63 g

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

brainly.com/question/4147359?referrer=searchResults

7 0

3 years ago

Chemical reactions constantly take place all around us—and even inside us. Many

Alika [10]

Answer:

I don't think you can go on without doing a reaction since you touch really anything so no you can not go without any chemical reaction I think this is the answer

5 0

3 years ago

Can someone help me plz

LenKa [72]

Answer:

See Explanation

Explanation:

The answer should be 8 mL according to my calculations!

7 0

4 years ago

In a five-fold serial dilution of a 20 ug/mL solution, all tube dilutions are 1/5. What is the substance concentration in the th

sergij07 [2.7K]

Answer:

In the third tube, the concentration is 0.16 ug/mL

Explanation:

In the first step, the solution is diluted by 5. Then, the concentration will be

20 ug/mL / 5 = 4 ug/mL

Then, in the second step this 4 ug / ml solution is diluted by a factor of five again:

4 ug /ml / 5 = 0.8 ug/mL

This solution is then diluted again by 5 and the concentration in the third tube will be then:

0.8 ug/mL / 5 = 0.16 ug/mL

Another way to calculate this is to divide the original concentration by the dilution factor ( 5 in this case) elevated to the number of dilutions. In this case:

Concentration in the third tube = 20 ug/mL / 5³ = 0.16 ug/mL

3 0

3 years ago