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3 years ago

Help needed ASAP, I will mark your answer as brainliest.

Chemistry

2 answers:

posledela3 years ago

5 0

B i think is the correct answer

evablogger [386]3 years ago

4 0

Answer:

Explanation:

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If body fats has a density of 0.94 g/mL and 3.0 liters of fat are removed, how many pounds of fat were removed from the patient?

-Dominant- [34]

Answer:

6.217 pounds

Explanation:

We are given;

Density of body fats 0.94 g/mL
Volume of fats removed = 3.0 L

We are required to determine the mass of fats removed in pounds.

We need to know that;

Density = Mass ÷ volume

1 L = 1000 mL, thus, volume is 3000 mL

Rearranging the formula;

Mass = Density × Volume

= 0.94 g/mL × 3000 mL

= 2,820 g

but, 1 pound = 453.592 g

Therefore;

Mass = 2,820 g ÷ 453.592 g per pound

= 6.217 pounds

Thus, the amount of fats removed is 6.217 pounds

4 0

3 years ago

A car rolls down a ramp. What is the force acting on the car that causes the movement down the ramp?

Komok [63]

The answer to your question is gravity. you are welcome. :-D

6 0

3 years ago

The solubility of a gas in a liquid increases with increasing pressure. To understand the above statement, consider a familiar e

weeeeeb [17]

Answer:

gukyfkuyk

Explanation:

5 0

3 years ago

Read 2 more answers

1. Complete the following

Marat540 [252]

Answer:

$\large \boxed{\text{0.603 mol}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 18.1

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}$

b) Moles of O₂

$\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}$

3 0

2 years ago

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

34kurt

Answer:

81 °C

Explanation:

This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:

q = heat added/released by a sample

m = mass of sample

c=specific heat of sample

ΔT = change in temperature

from here we can rearrange the equation to state:

q/(mc) = ΔT

1200J/((20.0g)(4.2J/g•°C)) = ΔT

14°C = ΔT

If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.

4 0

2 years ago