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harina [27]
3 years ago
7

Vanessa jogged 8 miles in 2 hours. What was her average speed?

Chemistry
2 answers:
sweet [91]3 years ago
5 0

\text{Hey there!}

\text{Vanessa jogged 8 miles in 2 hours. What was her average speed?}

\text{First,, highlight/ underline your key-terms. Then solve it from there!}

\text{Key term  1: \underline{8 miles}}\\\text{Key term 2: \underline{ in 2 hours}}

\text{Now we have to find her average speed. }

\text{Formula: }\dfrac{\text{a}}{\text{b}}

\text{a = 8}\\\text{b = 2}

\dfrac{8}{2}= \text{the answer}\\\\\dfrac{8\div2}{2\div2}=\dfrac{4}{1}= 4\\\\\\\boxed{\boxed{\bf{Answer: 4}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}\\\\\frak{LoveYourselfFirst:)}

rodikova [14]3 years ago
4 0
8miles in 2 hours. Let's find how many she did on average in one hour.

8/2 = 4miles per hour.
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Based on Chromium's position on the periodic table, which statement describes the element
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Answer:

C. chromium is a metal that is less reactive than sodium.

Explanation:

Hello.

Given the options:

A. chromium is a nonmetal and therefore a good conductor of heat and electricity .

B. chromium is a metal that is more reactive than potassium .

C. chromium is a metal that is less reactive than sodium .

D. chromium is a noble gas that is not reactive.

In this case, since chromium is in period 4 group VIB we infer it is a transition metal which slightly reacts with acids and poorly reacts with oxygen and other oxidizing substances. Thus, in comparison with both sodium and potassium which are highly reactive even with water as they get on fire, we can say that it is less reactive than both potassium and sodium, therefore, answer is: C. chromium is a metal that is less reactive than sodium.

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7 0
3 years ago
What is the molecular formula of a compound containing 89% cesium (Cs) and 11% oxygen (O) with a molar mass = 298 g/mol?
Maurinko [17]
M(Cs)=133 g/mol
M(O)=16 g/mol
M(CsxOy)=298 g/mol
w(Cs)=0.89
w(O)=0.11

CsxOy

x=M(CsxOy)w(Cs)/M(Cs)
x=298*0.89/133=2

y=M(CsxOy)w(O)/M(O)
y=298*0.11/16=2

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6 0
3 years ago
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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
3 years ago
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jekas [21]
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2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
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3) ΔG = ΔH - TΔS.
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3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
7 0
3 years ago
For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?
Stels [109]
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3 0
3 years ago
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