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4 years ago

Pablo drove 260 miles using 12 gallons of gas. at this rate, how many gallons of gas would he need to drive 286 miles?

1 answer:

7 0

260 miles with 12 gallons of gas
260 miles/12 gallons=21.6667 miles/1 gallon
286 miles/(21.6667 miles/1 gallon)=13.2 gallons of gas

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Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m

mina [271]

Answer:

<h3>I don't know what is the answer of your question sorry never mind..</h3>

Explanation:

<h3>And please marks me as brainliest... </h3>

3 0

3 years ago

Consider the following reaction at constant P. Use the information here to determine the value of Î”Ssurr at 355 K. Predict whet

Sloan [31]

The given reaction is as follows:

2NO (g) + O₂ (g) = 2NO₂ (g), ΔH = -114 kJ

It is known that dSsurr = -dHsys / T (Temp = 355 K)

So, dSsurr = - (-114 × 1000) / 355

dSsurr = +321.12 J/K

Hence, the value of dSsurr is +321.12 J/K

For a reaction to be spontaneous, dG<0,

Also dStotal = dSsys + dSsurr > 0

It is known that dG = dHsys - TdSsys,

Now let us assume,

dG<0

Also, dStotal = dSsys + dSsurr > 0

(-114 × 1000) - (355 × dSsys) <0

355 × dSsys > -114 × 1000

dSsys > -321

dSsys >dSsurr

dSsys + dSsurr > 0

dStotal > 0

Thus, the assumption is correct, and the given reaction is spontaneous. Hence, the final answer is Ssurr = +321 J/K reaction is spontaneous.

8 0

3 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago

What does the oxidation state of kr indicate

const2013 [10]

Kr look on periodic table it's krypton elements

8 0

4 years ago

I need help with this question

dsp73

Answer:

Explanation:

it just makes the most sense in my head

7 0

3 years ago

Read 2 more answers