Ask question

Ask question

All categories

3 years ago

9

Which is a balanced chemical equation? C7H16 + 5O2 6CO2 + 4H2O C7H16 + 11O2 7CO2 + 8H2O C7H16 + 14O2 7CO2 +5H2O C7H16 + 22O2 14C

O2 + 16H2O NextReset

2 answers:

pav-90 [236]3 years ago

6 0

<span>C7H16 + 11O2 ------> 7CO2 + 8H2O
</span><span>C- 7 C- 7
H- 16 H -8*2=16
O -11*2 =22 O - 7*2 +8 =14+8=22</span>

hram777 [196]3 years ago

6 0

Answer:

If you are a Plato user, it is B. C7H16 + 11O2 --> 7CO2 + 8H2O

Explanation:

You might be interested in

How might a strawberry plant depend on birds?

Helen [10]

Bird's eat the strawberries and they transport the seeds in their feces

8 0

3 years ago

Read 2 more answers

What type of element is phosphorus

Snezhnost [94]

I think the answer is non metal

6 0

3 years ago

Read 2 more answers

At sea level water usually boils at 100. °C and 760. mmHg pressure. On Mt Whitney, the pressure is about 560. mmHg. At what temp

WITCHER [35]

Answer:

91.7°C

Explanation:

We suppose you have a formula to work from. However, that is not supplied with this problem statement, so we looked one up.

The formula in the attachment is supposed to have good accuracy in the temperature range of interest. It gives vapor pressure of water in kPa, not mmHg, so we needed the conversion for that, too.

560 mmHg corresponds to about 74.66 kPa. The attached "Buck equation" formula is used to find the corresponding temperature. The exponential equation could be solved algebraically using logarithms and the quadratic formula, but we choose to find the solution graphically.

Water boils at about 91.7 °C on Mt. Whitney.

7 0

3 years ago

A solution is prepared by dissolving 33.0 milligrams of sodium chloride in 1000. L of water. Assume a final volume of 1000. lite

zheka24 [161]

Answer:

a. Molarity of NaCl solution = 5.64 * 10⁻⁷ mol/L

b. molarity of Na⁺ = 5.64 * 10⁻⁷ mol/L

c. molarity of Cl⁻ = 5.64 * 10⁻⁷ mol/L

d. Osmolarity = 1.128 osmol

e. mass percent of NaCl = 3.30 * 10⁻⁶ %

f. parts per million NaCl = 0.033 ppm NaCl

g. parts per billion of NaCl = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

Explanation:

Molarity of a solution = number of moles of solute (moles)/volume of solution (L)

where number of moles of solute = mass of solute (g)/molar mass of solute (g/mol)

a. Molarity of NaCl:

molar mass of NaCl = 58.5 g/mol, mass of NaCl = 33.0/1000) g = 0.033g

number of moles of NaCl = 0.033/58.5 = 0.000564 moles

Molarity of NaCl solution = 0.000564/1000 = 5.64 * 10⁻⁷ mol/L

b. Equation for the dissociation of NaCl in solution: NaCl ----> Na⁺ + Cl⁻

From the above equation I mole of NaCl dissociates to give 1 mole of Na⁺ ions,

Therefore molarity of Na⁺ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

c. From the above equation I mole of NaCl dissociates to give 1 mole of Cl⁻ ions,

therefore molarity of Cl⁻ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

d. From the above equation, dissociation of NaCl in water produces 1 mol Na⁺ and 1 mole Cl⁻.

Total number of particles produced = 2

Osmolarity of solution = number of particles * molarity of siolution

Osmolarity = 2 * 5.64 * 10⁻⁷ mol/L = 1.128 osmol

e. mass of percent of NaCl = {mass of NaCl (g)/ mass of solution (g)} * 100

density of water = 1 Kg/L

mass of water = 1 Kg/L * 1000 L = 1000 kg

1Kg = 1000 g

Therefore mass of solution in g = 1000 * 1000 = 1 * 10⁶ g

mass percent of NaCl = (0.033/1 * 10⁶) * 100 = 3.30 * 10⁻⁶ %

f. Parts per million of NaCl:

parts per million = 1 mg of solute/L of solution

One thousandth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.

Since the density of water is 1kg/L = 1,000,000 mg/L

1mg/L = 1mg/1,000,000mg or one part in one million.

parts per million NaCl = 33.0/1000 L = 0.033 ppm NaCl

g. Parts per billion = 1 µg/L of solution

1 g = 1000 µg

therefore, 33.0 mg = 33.0 * 1000 µg = 3.30 * 10⁴ µg

parts per billion of NaCl = 3.30 * 10⁴ µg/1000 L = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

5 0

3 years ago

Vitamin c is known chemically by the name ascorbic acid determine the empirical formula of ascorbic acid if it is composed of 40

kow [346]

Answer:

$=C_3H_4O_3$

Explanation:

When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,

Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C

Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H

Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O

Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.

Moles C = 3.41/3.41 = 1

Moles H = 4.58/3.41 = 1.34

Moles O = 3.41/3.41 = 1

Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3

Moles C = 1x3 = 3

Moles H = 1.34x3 = 4

Moles O = 1x3 = 3

Empirical Formula $=C_3H_4O_3$

3 0

3 years ago