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Strike441 [17]
2 years ago
8

Pharmacy technician measures the mass of a pill. The mass of one pill is

Chemistry
2 answers:
Galina-37 [17]2 years ago
8 0
The mass of 100 pills would be 78000 mg and 0.078 kilograms.

Hope this helps you :)
Murrr4er [49]2 years ago
6 0

Answer : The mass in kilograms of  100  pills is 0.078 kg.

Explanation :

First we have to convert the mass of one pill from 'mg' to 'kg'.

conversion used :

1mg=10^{-6}kg

So, 780mg=\frac{780mg}{1mg}\times 10^{-6}kg=780\times 10^{-6}kg

Now we have to calculate the mass for 100 pills.

As, 1 pills has mass = 780\times 10^{-6}kg

So, 100 pills has mass = 100\times 780\times 10^{-6}kg

                                     = 0.078 kg

Therefore, the mass in kilograms of  100  pills is 0.078 kg.

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A student runs an experiment in the lab and then uses the data to prepare an Arrhenius plot of the natural log of the rate const
Thepotemich [5.8K]

Answer:

21.86582KJ

Explanation:

The graphical form of the Arrhenius equation is shown on the image attached. Remember that in the Arrhenius equation, we plot the rate constant against the inverse of temperature. The slope of this graph is the activation energy and its y intercept is the frequency factor.

Applying the equation if a straight line, y=mx +c, and comparing the given equation with the graphical form of the Arrhenius equation shown in the image attached, we obtain the activation energy of the reaction as shown.

5 0
2 years ago
analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula
MrMuchimi
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr   
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr  /0.0135= 1 mol Cr
 0.0538 mol O/0.035= 4 mol Cr
K2CrO4
6 0
2 years ago
Read 2 more answers
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2
sashaice [31]

Answer:

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

(Ans)

ΔHf° of CaC2 = -59.0 kJ/mol

Explanation:

CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ

ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

Where

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

ΔHf°(CaC2) = -59.0 kJ/mol

7 0
2 years ago
Which of the following statements concerning hydrocarbons is/are correct?
Alona [7]

Answer:

1.  Saturated hydrocarbons may be cyclic or acyclic molecules.

2.  An unsaturated hydrocarbon molecule contains at least one double bond.

Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.

Therefore, first and second statements are correct.

Best regards.

6 0
2 years ago
Insoluble sulfide compounds are generally black in color. Which of the following combinations could yield a black precipitate? C
Lubov Fominskaja [6]

Answer:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

Explanation:

For these reactions, the products are:

1. Na₂S(aq) + 2 KCl(aq) → K₂S + 2 NaCl

2. Li₂S(aq) + Pb(NO₃)₂(aq) →  PbS + 2 Li(NO₃)

3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)

4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃

5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃

K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.

PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.

For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.

AgCl is an insoluble white compound.

As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!

Thus, combinations could yield a black precipitate are:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

I hope it helps!

6 0
2 years ago
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