Help??on the empty one

A 38-lb child is prescribed acyclovir for chicken pox in an amount of 80 mg/ kg body weight per day to be divided in four doses

Svetach [21]

Answer:

$\large \boxed{\text{Two tablets per day; half a tablet per dose}}$

Explanation:

1. Convert pounds to kilograms

$\text{Body weight} = \text{38 lb} \times \dfrac{\text{1 kg}}{\text{2.205 kg}} = \text{17.2 kg}$

2. Convert kilograms of body weight to milligrams of acyclovir

$\text{Mass of acyclovir} = \text{17.2 kg} \times \dfrac{\text{80 mg acyclovir}}{\text{1 kg}} = \text{1380 mg acyclovir }$

3. Convert milligrams of acyclovir to number of tablets

$\text{No. of tablets} = \text{1380 mg} \times \dfrac{\text{1 tablet}}{\text{700 mg}} = \textbf{2 tablets}$

4. Calculate the tablets per dose

The child must receive two tablets in four doses.

$\text{Tablets per dose} = \dfrac{\text{2 tablets}}{\text{4 doses}} = \textbf{0.5 tablets per dose}\\\\\text{The child must receive $\large \boxed{\textbf{two tablets per day and half a tablet per dose}}$}$

5 0

3 years ago

How many grams are 2.50 * 10^23 molecules of CaO?

jolli1 [7]

134.56 grams are 2.50 * 10^23 molecules of CaO.

Explanation:

According to Avagadro hypothesis, 1 mole of a substance contain 6.023 x 10^23 molecules or particles.

Data given:

Number of molecules of CaO =

FORMULA USED IN CALCULATION:

number of molecules = 6.02 x 10^23 x number of moles of CaO

At first number of moles is calculated:

$\frac{6.02 x 10^23}{2.50 X 10^23}$

= 2.4 moles of CaO is present

to calculate the mass of CaO in grams formula used is

mass = atomic mass x number of moles

atomic mass of CaO = 56.07 grams/mole

mass = 2.4 x 56.07

= 134.56 grams of calcium oxide is present is 2.50 * 10^23 molecules of CaO.

5 0

2 years ago

My recipe for chili requires a 125-g can of tomato paste

Digiron [165]

Answer:

here is where you can buy it

https://www.walmart.com/ip/Product-Of-Hunts-Tomato-Paste-Can-Count-1-Tomato-Paste-Grab-Varieties-Flavors/850453511?wmlspartner=wlpa&selectedSellerId=4648&&adid=22222222228000000000&wl0=&wl1=g&wl2=m&wl3=42423897272&wl4=aud-430887228898:pla-51320962143&wl5=9012391&wl6=&wl7=&wl8=&wl9=pla&wl10=115056687&wl11=online&wl12=850453511&veh=sem&gclid=Cj0KCQjw9ZGYBhCEARIsAEUXITUEOCgO6I5Xsg6DjNJSFFzf6pLMzFzhjW7WipqMInuuYo8EfjNoRAkaAp17EALw_wcB&gclsrc=aw.ds

4 0

1 year ago

Identify the conjugate base in each pairs

DiKsa [7]

Answer: 1) $RCOO^-$

2) $H_2PO_4^-$

3) $RNH_2$

4) $HCO_3^-$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

1) $RCOOH\rightarrow RCOO^-+H^+$

Here, $RCOOH$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $RCOO^-$ which is a conjugate base.

2) $H_3PO_4\rightarrow H_2PO_4^-+H^+$

Here, $H_3PO_4$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $H_2PO_4^-$ which is a conjugate base.

3) $RNH_3^+\rightarrow RNH_2+H^+$

Here, $RNH_3^+$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $RNH_2$ which is a conjugate base.

4) $H_2CO_3\rightarrow HCO_3^-+H^+$

Here, $H_2CO_3$ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms $HCO_3^-$ which is a conjugate base.

8 0

2 years ago

Write a balanced chemical equation for the incomplete combustion of methane to produce CO(g) as the only carbon-containing produ

Brums [2.3K]

Answer:

2CH4(g)+3O2(g)→2CO(g)+4H2O(g)

Explanation:

Start with methane (CH4) and the gas product CO. It is safe to assume that the remaining products and reactants are O2 and H2O. There we have an unbalanced equation CH4+O2->CO+H2O after which it can be balanced as above.

8 0

1 year ago