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1 year ago

14

The pressure in car tires is often measured in pounds per square inch (lb/in.2), with the recommended pressure being in the rang

e of 25 to 45 lb/in.2. Suppose a tire has a pressure of 38.5lb/in.2 . Convert 38.5lb/in.2 to its equivalent in atmospheres.
Express the pressure numerically in atmospheres.
38.5lb/in.2 = atm

1 answer:

Minchanka [31]1 year ago

4 0

Answer:

2.62 atm

Explanation:

1 atm = 14.6959 psi

38.5 / 14.6959 = 2.62 atm

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In the laboratory, a student dilutes 13.5 mL of a 11.6 M hydroiodic acid solution to a total volume of 100.0 mL. What is the con

igomit [66]

Answer: The concentration of the diluted solution is 1.566M.

Explanation:

The dilution equation is presented as this: $M_{s} V_{s} =M_{d} V_{d}$ .

·M= molarity (labeled as M)

·V= volume (labeled as L)

·s= stock solution (what you started with)

·d= diluted solution (what you have after)

Now that we know what each part of the formula symbolizes, we can plug in our data.

$11.6M*13.5mL=M_{d} *100mL$

We cannot leave it like this because the volumes must be in Liters, not milliliters. To convert this, we divide the milliliters by 1000.

$13.5mL/1000=0.0135L$ $100mL/1000=0.1L$

Now that we have the conversions, let's plug them into the equation.

$11.6M*0.0135L=M_{d} *0.1L$

The only thing that we need to do now is actually solving the answer.

$M_{d} =\frac{11.6M*0.0135L}{0.1L}$ $M_{d} =1.566M$

From the work shown above, the answer is 1.566M.

I hope this helps!! Pls mark brainliest :)

7 0

1 year ago

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what

melomori [17]

Answer:

$Y=65.7\%$

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

$n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2$

Thus, we compute the percent yield as:

$Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%$

Best regards.

8 0

2 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

$Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}$

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

6 0

3 years ago

2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??

Zepler [3.9K]

Answer:

$d=1.78\times 10^{-7}\ Dg/\mu m^3$

Explanation:

Given,

The density of Helium is $1.78\times 10^4\ g/cm^3$

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

$d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3$

So, the density of Helium is equal to $1.78\times 10^{-7}\ Dg/\mu m^3$ .

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2 years ago

What statement correctly describes gravity

Gre4nikov [31]

Answer:

gravitation

Explanation:

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2 years ago

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