Burning coal releases the following pollutants except

When you see the word heptahydrate, what number should be written in front of the formula for water? 5 6 7 8?

Yuki888 [10]

Hept for 7, hence the number is seven.

8 0

3 years ago

Evaluate the exponential expression (−2)6.

Olegator [25]

A general exponential expression is something like:

A^n

This means that we need to multiply the number A by itself n times.

Using that we will get (-2)^6 = 64

With that definition, we can rewrite:

(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)

So we just need to solve the above expression.

Also, remember the rule of signs:

(-)*(-) = (+)

We will get:

(-2)*(-2)*(-2)*(-2)*(-2)*(-2) = [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]

= 4*4*4 = 16*4 = 64

Then we got:

(-2)^6 = 64

If you want to learn more, you can read:

brainly.com/question/17172630

6 0

2 years ago

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)<br><br> Which type of reaction is this?

erastova [34]

Answer: Combustion

Explanation:

3 0

1 year ago

Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa

kumpel [21]

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

$Y_{O_{2} } = \frac{(P_{O_{2} })^{h} } {(P_{50})^{h} + (P_{O_{2} })^{h} }$

Where $Y_{O_{2} } \\$ is the fractional oxygen saturation

${P_{O_{2} }$ is the partial pressure of oxygen

$P_{50}$ is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

${P_{O_{2} }$ = 40 mm Hg

$P_{50}$ = 22 mm Hg

h = 3

Putting these values into the equation, we get

$Y_{O_{2} } = \frac{(P_{O_{2} })^{h} } {(P_{50})^{h} + (P_{O_{2} })^{h} }$

$Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3} }$

$Y_{O_{2} } = \frac{64000 }{10648 + 64000 }$

$Y_{O_{2} } = \frac{64000 }{74648 }$

$Y_{O_{2} } = 0.86$

Hence, the fractional saturation for hemoglobin is 0.86.

4 0

3 years ago

You placed a sample of a hydrate of calcium chloride (CaCl2) in a weighed test tube, and weighed the filled test tube.

ziro4ka [17]

Answer:

1. 5g

2. 2.3g

3. 2.7g

4. 0.02mol

5. 0.13mol

6. 7moles

Explanation:

From the question, the following were obtained:

Mass of empty tube = 13.5g

Mass of empty tube + hydrated salt = 18.5g

Mass of tube + anhydrous salt = 16.2g

1. Mass of empty tube = 13.5g

Mass of empty tube + hydrated salt = 18.5g

Mass of hydrated salt = 18.5 — Mass of empty tube

Mass of hydrated salt = 18.5 — 13.5 = 5g

2. Let us calculate the mass of the anhydrous salt.

Mass of tube + anhydrous salt = 16.2g

Mass of empty tube = 13.5g

Mass of anhydrous salt = 16.2 — Mass of empty tube = 16.2 — 13.5

Mass of anhydrous salt = 2.7g

Now we can calculate the mass of the water evolved as follows:

Mass of water = Mass of hydrated salt — Mass of anhydrous

Mass of water = 5 — 2.7 = 2.3g

3. Mass of empty tube = 13.5g

Mass of anhydrous salt = 16.2 — Mass of empty tube = 16.2 — 13.5

Mass of anhydrous salt = 2.7g

4. MM of CaCl2 = 40 +(2x35.5) = 40 + 71 = 111g

Mass of CaCl2 = 2.7g

Number of mole = Mass /Molar Mass

Number of mole of CaCl2 = 2.7/111 = 0.02mol

5. MM of H2O = (2x1) +16 = 2 + 16 = 18g/mol

Mass of H2O = 2.3g

Number of mole = Mass /Molar Mass

Number of mole of H20 = 2.3/18 = 0.13mol

6. To get the mole of water in the molecular formula, we will find the ratio of the number of mole of anhydrous salt to water as shown below:

Mole anhydrous : mole of water ie

0.02 : 0.13 = 1 : 7

Therefore, the mole of water in the formula is 7 ie

CaCl2.7H20

3 0

3 years ago