How do you round 431,801 kg to four significant figures

A very hot cube of copper metal (32.5 g) is submerged into 105.3 g of water at 15.4 0C and it reach a thermal equilibrium of 17.

zysi [14]

Answer:

The initial temperature of the metal is 84.149 °C.

Explanation:

The heat lost by the metal will be equivalent to the heat gain by the water.

- (msΔT)metal = (msΔT)water

-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C

-ΔT = 836.29/12.51 °C

-ΔT = 66.89 °C

-(T final - T initial) = 66.89 °C

T initial = 66.89 °C + T final

T initial = 66.89 °C + 17.3 °C

T initial = 84.149 °C.

7 0

3 years ago

When the pressure on 2.5 liters of oxygen is decreased from 2.0 atm to 1.0 atm, the volume of the gas

uranmaximum [27]

Answer:

volume of the gas is 5.0L

Explanation:

Using Boyle's law that state the pressure of a gas is inversely proportional to volume of it occupies when temperature is constant, it is possible to write:

P₁V₁ = P₂V₂

Where P is pressure, V is volume and 1 and 2 are initial and final states.

If initial volume is 2.5L, initial pressure is 2.0atm and 1.0atm is final pressure, final volume is:

2.0atm*2.5L = 1atm V₂

5.0L = V₂

Thus, volume of the gas is 5.0L.

5 0

3 years ago

Combustion of hydrocarbons such as undecane (C_11H_24) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Eart

bekas [8.4K]

Answer:

Volume of carbon dioxide is 428.23 L.

Explanation:

Below is the chemical reaction or chemical equation for the combustion of hydrocarbons such as undecane into carbon dioxide.

$C_{11}H_{24}(l) + 17O_{2}(g) \to 11CO_{2} (g) + 12H_2O(g)$

Here, undecane is in liquid form that reacts with gaseous oxygen (combustion) and produces carbon dioxide and water as a product in the gaseous form.

The molar mass of undecane = $156.31 g/mol.$

$\text{Number of moles} = \frac{260g}{156.31g/mol} = 1.66 \ mol.$

From the equation, it can be seen that 1 mole of undecane produces 11 moles of carbon dioxide. Therefore, 1.66 mol will produce 18.26 mol of carbon dioxide.

Now find the volume of 18.26 mol of carbon dioxide when the temperature is 13 degrees Celsius and pressure is 1 atm.

$V = \frac{nRT}{P} \\$