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3 years ago

7

Calculate the number of joules needed to change the temperature of 40.0 g of water from 33.0 degree c to 43.0 degree c

1 answer:

nikdorinn [45]3 years ago

7 0

Answer:

c

Explanation:

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At a certain temperature the rate of this reaction is first order in SO₃ with a rate constant of 0.208 s⁻¹:

NNADVOKAT [17]

Answer:

0.30 M

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t = ?

$[A_0]$ is the initial concentration = 1.36 M

k is the rate constant = 0.208 s⁻¹

t = 7.30 seconds

So,

$[A_t]=1.36\times e^{-0.208\times 7.30}\ M=0.30\ M$

8 0

3 years ago

Calculate the concentration of the following solution in mol/dm3 0.1 moles of NaCl in 200 cm3

taurus [48]

1 $cm ^{3}$ = 0.001 $dm ^{3}$ . Therefore 200 $cm ^{3}$ = 0.2 $dm ^{3}$ . Molarity = $\frac{number of moles of NaCl}{volume of the solution} = \frac{0.1}{0.2} = 0.5 mol/dm^{3}$ . Hope this helps.

8 0

3 years ago

What is the formula for sulfur hexa fluoride?

vlabodo [156]

S2F6, is the correct answer I believe!

6 0

3 years ago

Read 2 more answers

Waves transport water in the deep ocean toward the coastlines. True or false

ASHA 777 [7]

Answer:

Most likely true

Explanation:

because the speed of all ocean waves is controlled by gravity, wavelength, and water depth. ... Waves moving through water deeper than half their wavelength are known as deep-water waves. On the other hand, the orbits of water molecules in waves moving through shallow water are flattened by the proximity of the sea surface bottom.

7 0

3 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

3 years ago