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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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Suspensions

Explanation:

Suspensions are heterogeneous mixtures that contains large particles that can settle out or be filtered.

  • Suspensions are mixtures of small insoluble particles of a solid in a liquid or gas.
  • Examples are:
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  2. muddy water
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The particles in suspension can settle on standing

Learn more:

Suspension brainly.com/question/1557970

heterogeneous mixture brainly.com/question/1446244

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8 0
3 years ago
Which of the following describes an impact of the specific heat of water on the planet and the life forms that exist on it?
photoshop1234 [79]

Answer: D) sweat requires a lot of thermal energy to be absorbed to evaporate

Explanation:

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8 0
2 years ago
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A certain liquid has a density of 0.70g/m. Find the mass of 6.4L o this liquid
neonofarm [45]

The mass of the liquid is 4.8 kg.

6.4 L = 6400 mL

Mass = 6400 mL × (0.70 g/1 mL) = 4500 g = 4.8 kg

5 0
3 years ago
At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall
mestny [16]

Answer:

concentration of [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 0.0248 = 2.48 ×10⁻² M

concentration of [CO_2] = 0.4442 M

Explanation:

Equation for the reaction:

2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Concentration of   CO_2_{(g) = \frac{2.3}{4.9}  = 0.469

For our ICE Table; we have:

                       2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Initial                 0.469                              0                           0

Change              - 2x                                +2x                      +x

Equilibrium       (0.469-2x)                       2x                         x

K = \frac{[CO]^2[O]}{[CO_2]^2}

K = \frac{[2x]^2[x]}{[0.469-2x]^2}

4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}

Since the value pf K is very small, only little small of  reactant goes into product; so (0.469-2x)² = (0.469)²

4.1*10^{-6} = \frac{2x^3}{(0.938)}

2x^3 =3.8458*10^{-6

x^3 =\frac{3.8458*10^{-6}}{2}

x^3=1.9229*10^{-6

x=\sqrt[3]{1.9929*10^{-6}}

x = 0.0124

∴ at equilibrium; concentration of  [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 2x  = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of [CO_2] = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

3 0
3 years ago
Calculate the volume in dm3 occupied by 26.4 g of carbon dioxide​
Mama L [17]
  • Density of carbon dioxide=1564kg/m^3
  • Mass=26.4g=0.0264kg

\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}

\\ \sf\longmapsto Volume=\dfrac{Mass}{Density}

\\ \sf\longmapsto Volume=\dfrac{0.0264}{1564}

\\ \sf\longmapsto Volume=1.687m^3=1.69m^3

  • 1m^3=1000dm^3

\\ \sf\longmapsto Volume=1.69(1000)=1690dm^3

7 0
2 years ago
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