Ask question

Ask question

All categories

3 years ago

7

Joel has noticed that his engine is not compressing properly during the compression and ignition step. What problem will this po

se?

1 answer:

denis-greek [22]3 years ago

7 0

The answer is: The engine will run inefficiently APEX....

You might be interested in

In a separate experiment beginning from the same initial conditions, including a temperature Ti of 25.0°C, half the number of mo

NNADVOKAT [17]

Answer:

- 178 ºC

Explanation:

The ideal gas law states that :

PV = nRT,

where P is the pressure, V is the volume, n is number of moles , R is the gas constant and T is the absolute temperature.

For the initial conditions :

P₁ V₁ = n₁ R T₁ (1)

and for the final conditions:

P₂V₂= n₂ R T₂ where n₂ = n₁/2 then P₂ V₂ = n₁/2 T₂ (2)

Assuming V₂ = V₁ and dividing (2) by Eqn (1) :

P₂ V₂ = n₁/2 R T₂ / ( n₁ R T₁) then P₂ / P₁ = 1/2 T₂ / T₁

4.10 atm / 25.7 atm = 1/2 T₂ / 298 K ⇒ T₂ = 0.16 x 298 x 2 = 95.1 K

T₂ = 95 - 273 = - 178 º C

6 0

3 years ago

lana66690 [7]

Answer:

c

Explanation:

3 0

3 years ago

Read 2 more answers

A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what

Charra [1.4K]

1) mass composition

N: 30.45%
O: 69.55%
-----------
100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

=> N2O4

Answer: N2O4

3 0

2 years ago

Suppose a friend says that we don't need to worry about the rising temperatures associated with global climate change. Her claim

ludmilkaskok [199]

Answer:

Carbonic acid could be formed.

Explanation:

Hello,

Based on her claim, it would be a really useful strategy to prevent global warming, nevertheless, there would be a problem if a increasing amount of carbon dioxide is not buried at the bottom of the ocean yet it flows freely along the sea and probably reacting with the water, causing carbonic acid to be formed and subsequently cutting back the sea's pH (increasing its acidity).

It would be useful, but a constant monitoring of the sea's pH must be needed because this could cause some species to be affected not only by the temperature but for the acid pH as well.

Best regards.

4 0

2 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

<u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

<u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago