All categories

3 years ago

these healthcare professionals are usually limited to the types of prescriptions they can write by a set standing orders.

Chemistry

1 answer:

Norma-Jean [14]3 years ago

5 0

I'm not sure what your question is?

You might be interested in

Can you check this for me?

Aliun [14]

Hi, your answer is correct.

3 0

3 years ago

The energy required to break existing chemical bonds in reactants is called the ______ energy.

mixer [17]

The energy required to break existing chemical bonds in reactants is called the activation energy.

<h3>What is activation energy?</h3>

Activation energy in chemistry is the energy required to initiate a chemical reaction.

Chemical reactions involve the breaking of chemical bonds in substances called reactants to form new substances called products.

The energy required to break the bond in the existing reactants thus elevating these substances to a state of high activation is known as activation energy.

Therefore, it can be said that energy required to break existing chemical bonds in reactants is called the activation energy.

Learn more about activation energy at: brainly.com/question/11334504

#SPJ1

8 0

1 year ago

The medium for ocean waves

stepladder [879]

This is a tough one but the answer is water

think of it this way the ocean is water

7 0

2 years ago

MgNH4PO4•6H2O loses H2O in a stepwise fashion as it is heated. Between 40 degrees celsius and 60 degrees celsius, the monohydrat

tino4ka555 [31]

Answer:

$\%\ Composition\ of\ phosphorus=20.00\ \%$

Explanation:

The molecular formula of the monohydrate formed = $MgNH_4PO_4.H_2O$

The molecular mass of the monohydrate formed = $Mass_{Mg}+Mass_{N}+6\times Mass_{H}+5\times Mass_O$

So, Mass = 24 + 14 + 6 × 1 + 5 × 16 = 155 g

Mass of phosphorus = 31 g

Thus,

$\%\ Composition\ of\ phosphorus =\frac{Mass_{phosphorus}}{Total\ mass}\times 100$

$\%\ Composition\ of\ phosphorus=\frac{31}{155}\times 100$

$\%\ Composition\ of\ phosphorus=20.00\ \%$

4 0

3 years ago

12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?

Kaylis [27]

Answer:

$115.625^{\circ}\text{F}$

Explanation:

$m_1$ = First mass of water = 12 oz

$m_2$ = Second mass of water = 20 oz

$\Delta T_1$ = Temperature difference of the solution with respect to the first mass of water = $(T-75)^{\circ}\text{F}$

$\Delta T_2$ = Temperature difference of the solution with respect to the second mass of water = $(T-75)^{\circ}\text{F}$

c = Specific heat of water

As heat gain and loss in the system is equal we have

$m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}$

The final temperature of the solution is $115.625^{\circ}\text{F}$ .

3 0

2 years ago