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GarryVolchara [31]
2 years ago
8

4. Traditional bomb calorimetry can be used to find the energy content of food. Why is it

Chemistry
1 answer:
VladimirAG [237]2 years ago
3 0

Answer:

B. it accounts for all the energy in the good even if some of its largely excreted by th body

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A rift valley forms at a
vovikov84 [41]
D is the answer your looking for
4 0
3 years ago
An I.V. infusion order is received for Lidocaine 2 g in 250 ml of 0.9% NaCl to be infused at 3 mg/min. What will the flow rate b
Tamiku [17]

Answer:

The flow rate would be 22.5ml/hr

Explanation:

Volumetric flow rate = Mass flow rate ÷ density

Mass flow rate = 3mg/min = 3mg/min × 60min/1hr = 180mg/hr

Density = mass/volume = 2g/250ml = 0.008g/ml = 0.008g/ml × 1000mg/1g = 8mg/ml

Volumetric flow rate = 180mg/hr ÷ 8mg/ml = 22.5ml/hr

5 0
3 years ago
Which of the following does not contribute to water vapour to the atmosphere
vampirchik [111]

Answer:

<h2>Industry </h2>

Explanation:

because industry is a building not a weather or related on nature

3 0
2 years ago
What reagent is <br>used to test for starch<br>​
Elodia [21]

Answer:

iodine

Explanation:

In the presence of starch, iodine turns a blue/black colour. It is possible to distinguish starch from glucose (and other carbohydrates) using this iodine solution test. For example, if iodine is added to a peeled potato then it will turn black. Benedict's reagent can be used to test for glucose.

8 0
3 years ago
What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?
soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

P

V

=

n

R

T

a

a

∣

∣

−−−−−−−−−−−−−−−

, where

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

M

M

=

m

n

a

a

∣

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

ρ

=

m

V

a

a

∣

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

M

⋅

R

T

Rearrange to get

P

V

⋅

M

M

=

m

⋅

R

T

P

⋅

M

M

=

m

V

⋅

R

T

M

M

=

m

V

⋅

R

T

P

Finally, use equation

(

2

)

to write

M

M

=

ρ

⋅

R

T

P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M

M

=

1.02

g

L

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

273.15

+

37

)

K

0.990

atm

M

M

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

26.3 g mol

−

1

a

a

∣

∣

−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three

7 0
3 years ago
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