A student lifts a box of books that weighs 350 N. The box is lifted 4.0 m. How much work does the student do on the box?

1 Calcium carbonate breaks down on heating to produce calcium oxide and carbon dioxide gas.

Alborosie

Answer: See below

Explanation:

1. a) 0.15 moles calcium carbonate (15g/100g/mole)

b) 0.15 moles CaO (molar ratio of CaO to CaCO3 is 1:1)

c) 8.4 grams CaO (0.15 moles)*(56 grams/mole)

2. a) 0.274 moles Na2O (17g/62 grams/mole)

b) 46.6 grams NaNO3 (2 moles NaNO3/1 mole Na2O)*(0.274 moles Na2O)*(85 g/mole NaNO3)

7 0

3 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion.

Alina [70]

Answer: The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_4^{2-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2$

Dividing 2 from 3, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2$

3 0

3 years ago

What is the mass in grams of 5.00moles of CH4?

anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0

3 years ago

What is the percent of O in CO2?

katen-ka-za [31]

Answer:

72.7% of oxygen

Explanation:

8 0

2 years ago

A water bath in a physical chemistry lab is 1.85m long, 0.810m wide and 0.740m deep. If it is filled to within 2.57 in from the

jenyasd209 [6]

Answer:

The volume of water in water bath is 1,011 Liters.

Explanation:

Length of the water bath, L = 1.85 m

Width of the water bath, W= 0.810 m

Height of the water bath ,H= 0.740 m

Height of the water in water bath, h= 0.740 m - 2.57 inches

1 m = 39.37 inch

$= 0.740 m - \frac{2.57}{39.37} m = 0.6747 m$

Volume of the water in bath = L × W × h

$=1.85 m\times 0.810 m\times 0.6747 m=1.011 m^3$

$1 m^3=1000 L$

$1.011 m^3=1.011\times 1000 L=1,011 L$

The volume of water in water bath is 1,011 Liters.

6 0

3 years ago