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3 years ago

A patient needs to take 875mg twice daily The pills in the bottle are each 250mg How many pills does she wants to take 875mg

Chemistry

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer:

$3.5$ pill need to taken for completing dosage of $875$ mg

Explanation:

Total dosage of medicine to be taken in a day

$= 2 * 875 \\= 1750$ mg

Total weight of one pill $= 250$ mg

Number of pills required is equal to total dosage of medicine divided by the weight of one pill.

Hence,

Number of pills

$= \frac{875}{250} \\= \frac{175}{50} \\= \frac{35}{10} \\= 3.5$

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

If nitrogen and hydrogen combine in a combustion reaction, what would the product of the reaction be?

kow [346]

Co2

Explanation:

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

4 0

3 years ago

Read 2 more answers

A solution is classified as a weak base. Which of these could be the pH of the solution?

Pepsi [2]

Explanation:

The pH of a weak base falls somewhere between 7 and 10.

3 0

3 years ago

When someone opens a bottle of carbonated drinks, the pressure inside the bottle decreases.

Mekhanik [1.2K]

Answer:

When you open the bottle, there is a dramatic decrease in pressure over the liquid, so the CO2 starts to leave the liquid very rapidly, causing the mass exodus of gas, or "explosion" of bubbles. This also explains why soda goes flat.

Explanation: HOPE THIS HELPS YOU..

8 0

3 years ago

Calculate the amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C.

qwelly [4]

Answer:

The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

Between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

c= 4.184 $\frac{J}{g*C}$
m= 32 g
ΔT= Tfinal - Tinitial= 22°C - 8°C= 14°C

Replacing:

Q= 32 g* 4.184 $\frac{J}{g*C}$ *14 °C

Solving:

Q= 1,874.432 J

The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J

7 0

2 years ago