Highlight the claim.

In which of the following compounds is the mass ratio of chromium to oxygen closest to 1.62 to 1.00?

ohaa [14]

Answer:

The answer to your question is letter A

Explanation:

Data

Mass ratio 1.62 to 1.00

Atomic mass Cr = 52 g

Atomic mass O = 16

Process

1.- Calculate the proportion Chromium to Oxygen to each possible solutions

a) CrO₃ = $\frac{52}{16 x 3} = \frac{52}{48} = 1.08$

b) CrO₂ = $\frac{52}{16 x 2} = \frac{52}{32} = 1.63$

c) CrO = $\frac{52}{16} = 3.25$

d) Cr₂O = $\frac{52 x 2}{16} = \frac{104}{16} = 6.5$

8 0

3 years ago

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

3 years ago

2 C4H10 + 13 O2--> 8 CO2 + 10 H2O

AleksAgata [21]

Answer:

4.14 x 10²⁴ molecules CO₂

Explanation:

2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O

To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.

4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀

100 grams C₄H₁₀ 1 mol C₄H₁₀ 8 mol CO₂
-------------------------- x ---------------------- x ---------------------
58.124 g 2 mol C₄H₁₀

6.022 x 10²³ molecules
x ------------------------------------ = 4.14 x 10²⁴ molecules CO₂
1 mol CO₂

7 0

2 years ago

Which greenhouse gas typically exists in both the stratosphere and the troposphere?

Arturiano [62]

The answer should be ozone

5 0

3 years ago

Read 2 more answers

After being thoroughly stirred at 10.°C, which mixture is heterogenous?(1) 25.0 g of KCl and 100. g of H2O

zheka24 [161]

The answer is (2) KNO3. This depends on the solubility of these four compounds at 10℃. For NaCl, it is 35.8 g, For NaNO3, 80.8 g. KCl, 31.2 g. KNO3, 21.9g. So only KNO3 is less than 25.0 g.

5 0

3 years ago