# What does tying the arm above the elbow do?

Answer: Tying a piece of rope or string above your elbow makes your vein pop out making it easier to a insert aneedle

## Related Questions

Find the density of an object that has a mass of 66g and a 10 cm3. HELP ASAP

Explanation:

The formula to calculate the density is given by :-

We are given that the volume of an object is

Mass of object  = 66g

Now, the density of object is given by :-

Hence, the density  of object =

Density = mass /volume

Density = 66g/10cm³

Density = 6.6g/cm³....

Just take note of the formula....

Hope this helped....

The mass of one gram A- is kept standard platinum cylinder in france
B- is .01 kg
C-equals the mass of 1ml of water at 4 degrees celsius
D- equals 1 cm^3

The answer is C. equals the mass of 1 mL of water at 4°C because the density of water = 1 g / mL at 4 C . So the mass of 1 mL of water will be equal to 1 gram.

N order is written for 10 ml of a 0.125% bupivacaine injection and 10 ml of morphine injection (1 mg/ml) to be mixed with 250 ml of normal saline. the infusion is to be administered over 3 hours. the iv set delivers 20 drops/ml. what should be the rate of flow in drops/minute to deliver this infusion?

The total volume of the infusion in 10 mL (bupivacaine) + 10 mL (morphine) + 250 mL (saline solution) = 270 mL.
The number of drops is: (270 mL)*(20 drops/mL) = 5400 drops
If this is to be administered over 3 hours, or over 180 minutes, then the number of drops per minute is:
5400 drops / 180 minutes = 30 drops/minute

A 13.5 g sample of gold is heated to 125.0°C, then placed in a calorimeter containing 60.0 g of water. The final temperature of the water 20.00 o C. The specific heat of gold is 0.130 J/g o C. What was the initial temperature of the water?

The initial temperature was 19.27 °C.

The guiding principle is the Law of Conservation of Energy: the sum of all the energy transfers must add up to zero.

The formula for the heat q gained or lost by a substance is

q = mCΔT

where

m = the mass of the substance.

C = its specific heat capacity.

ΔT = T_f - T_i = the change in temperature.

In this problem, there are two heat transfers.

Heat lost by gold + heat gained by water = 0

m _1C_1ΔT_1 + m_2c_2ΔT_2= 0

m_1 = 13.5 g; C_1 = 0.130 J·°C^(-1)g^(-1); ΔT_1 = T_f – T_i = 20.00 °C – 125.0 °C = -105.0 °C

m_2 = 60.0 g; C_2 = 4.184 J·°C^(-1)g^(-1); ΔT_2 = ?

q_1 = m_1C_1ΔT_1 = 13.5 g × 0.130 J·°C^(-1)g^(-1) × -105.0 °C = -184.3 J

q_2 = m_2C_2ΔT_2 = 60.0 g × 4.184 J·°C^(-1)g^(-1) × ΔT_2

= 251.0 ΔT_2 J·°C^(-1)

q_1+ q_2 = -184.3 J + 251.0 ΔT_2 J·°C^(-1) = 0

251.0 ΔT_2 °C^(-1) = 184.3

ΔT_2 = 184.3/251.0 °C^(-1) = 0.734°C

ΔT_2 = T_f - T_i = 20.00 °C -T_i = 0.734 °C

T_i = 20.00 °C – 0.734 °C = 19.27 °C

Compare van der waals forces ionic bonds and covalent bonds

1) Covalent bond: The bond that is formed by mutual sharing of electrons.

2) Ionic bond: The bond that is formed by complete transfer of electron from one atom to an other atom.

3) Van der waal: These are weak interactions between one molecules with other polar or non polar molecules to hold to each other.

Explanation:

a) Similarities between van der waal and ionic bonds:

As both are considered as a inter-molecular forces, however ionic bonds are stronger than van der walls forces.

b) Difference between van der waal and ionic bonds:

As ionic bonds are formed by complete transfer of electrons from one atom to an other atom while in van der waal, a slight attraction is develop when oppositely charged regions molecules come close to each others.

c) Difference between ionic and covalent bonds:

As both are strong bond, but if we compare both, the ionic bond are stronger than covalent bond.

Covalent bonds are bonds formed between two non-metals, while Ionic bonds are formed between a metal and a non-metal.

Waves we cannot actually see (unlike ripples) and those not needing a medium to travel within belong to this category of waves?

Well I don't know for sure but it can be sound waves.

The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2: C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(g)

a) How many moles of CO2 are produced when 0.300 mol of C6H12O6 reacts in this fashion?

b) How many grams of C6H12O6 are needed to form 2.00 g of C2H5OH?

c) How many grams of CO2 form when 2.00 g of C2H5OH are produced?

A)  0.300 mol of C6H12O6...

0.300 mol C6H12O6 * (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2

B) 2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol

2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

C) MM of CO2 = 44.01 g/mol

MM of C2H5OH = 46.07 g/mol

2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

What are the products of the fermentation of glucose?

The fermentation products from glucose were butyrate, acetate, and formate; those from lactate were 1-butanol, ethanol, butyrate, and formate.

What causes glucose to ferment?

When yeast is added it feeds on the sugar in the absence of oxygen to form wine (a solution of ethanol) and carbon dioxide. A chemical reaction called fermentation takes place in which the glucose is broken down to ethanol by the action of enzymes in the yeast.

#SPJ2

To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.

A) First start out with the 0.300 mol of C6H12O6...
0.300 mol C6H12O6 * (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2

*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*

B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
This can be used as a conversion factor to change grams to moles.
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH * (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6

Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
0.0217 mol C6H12O6 * (180.16 g C6H12O6 / 1 mol C6H12O6) = 3.91 g C6H12O6

C) Now I am going to put it all into one long dimensional analysis problem.
MM of CO2 = 44.01 g/mol
MM of C2H5OH = 46.07 g/mol

2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!

What does a light bulb, a toaster, a radio, and a computer all have in common??? Help with science please...

We have that the above mentioned device all have the concept of radiation in common

From the question we are told

What does a light bulb, a toaster, a radio, and a computer all have in common???

Generally

it is important to not that Radiation can be  defined as the emission of energy from a body in the form of a wave.

Therefore

We can see that with all of the above mentioned device are device that radiate one form of wave or particle

light bulb -Heat and Light waves

• A toaster- Heat Waves
• Therefore we can say that

The above mentioned device all have the concept of radiation in common