When combined in the correct ratio hydrogen and oxygen atoms can form water as shown below

Which of the following conditions creates the largest waves?

SashulF [63]

Answer:

strong winds that blow in short spurts over a small distance

Explanation:

Generally, the most common factor that causes the largest waves in the ocean is winds. this is also called wind-driven waves of the ocean when the surface wind flows on the surface of the ocean, this disturbance creates the friction between surface wind and surface of the ocean which creates waves in the ocean

6 0

3 years ago

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Iron has a density of 7.9 g/cm3. The volume of a regular cylinder is V =πr2h. An iron cylinder has a height of 3.00 m and a mass

Sedaia [141]

Hfhsnebrjicjcbdnd hope this helps

6 0

2 years ago

15.0 g of cream at 10.0 ℃ are added to an insulated cup containing 150.0 g of coffee at 78.6 °C. Calculate the equilibrium tempe

elena-14-01-66 [18.8K]

Answer:

The equilibrium temperature of the coffee is 72.4 °C

Explanation:

Step 1: Data given

Mass of cream = 15.0 grams

Temperature of the cream = 10.0°C

Mass of the coffee = 150.0 grams

Temperature of the coffee = 78.6 °C

C = respective specific heat of the substances( same as water) = 4.184 J/g°C

Step 2: Calculate the equilibrium temperature

m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)

15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)

62.76T2 - 627.6 = -627.6T2 + 49329.36

690.36T2 = 49956.96

T2 = 72.4 °C

The equilibrium temperature of the coffee is 72.4 °C

7 0

3 years ago

Based on the equation, how many grams of Br2 are required to react completely with 72.4 grams of AlCl3

svlad2 [7]

Answer: 131 g of bromine is required.

Explanation:

The balanced equation will be :

$2AlCl_3+3Br_2\rightarrow 2AlBr_3+3Cl_2$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $AlCl_3$

$\text{Number of moles}=\frac{72.4g}{133g/mol}=0.544moles$

According to stoichiometry :

2 moles of $AlCl_3$ require = 3 moles of $Br_2$

Thus 0.544 moles of $AlCl_3$ require= $\frac{3}{2}\times 0.544=0.816moles$ of $Br_2$

Mass of $Br_2=moles\times {\text {Molar mass}}=0.816moles\times 160g/mol=131g$

Thus 131 g of bromine is required.

5 0

2 years ago

An ideal gas at a given initial state expands to a fixed final volume. would the work be greater if the expansion occurs at cons

crimeas [40]

Answer:

Constant pressure

Explanation:

At constant pressure,

$w = -p\Delta V = -p(V_{f} - V_{i})$

At constant temperature,

$w = -RT \ln \left(\dfrac{V_{f}}{V_{i}} \right)$

1 mol of an ideal gas at STP has a volume of 22.71 L.

Let's compare the work done as it expands under each condition from an initial volume of 22.71 L.

Isobaric expansion

$w = -100p(V_{2} - 22.71}); \text{(1 bar$\cdot$L = 100 J)}$

A plot of -w vs V₂ gives a straight line (red) with a constant slope of 100 J/L as in the diagram below (Note that w is work done on the system, so -w is the work done by the system). \

Isothermal expansion

$w= -8.314 \times 273.15 \ln \left(\dfrac{V_{f}}{22.71} \right)\\\\= -2271 \left( \ln V_{f} -\ln22.71 \right)\\= -2271 \left(\ln V_{f} - 3.123 \right)\\= 7092 - 2271\ln V_{f}$

A plot of -w vs V₂ is a logarithmic curve. Its slope starts at 100 J/mol but decreases as the volume increases (the blue curve below).

Thus, the work done during an expansion at constant pressure is greater than if the system is at constant temperature.

4 0

3 years ago