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3 years ago

PLEASE HELP ASAP!!!!

Chemistry

1 answer:

Vsevolod [243]3 years ago

4 0

Answer:

The answer to your question is C = 0.037 cal/g°C

Explanation:

Data

mass of the wire = m = 237 g

temperature 1 = T1 = 25°C

temperature 2 = T2 = 107°C

Heat = Q = 722 cal

Specific heat = C

Process

1.- Write the formula to find the specific heat

Q = mC(T2 - T1)

-Solve for C

C = Q / m(T2 - T1)

2.- Substitution

C = 722 / 237(107 - 25)

3.- Simplification

C = 722 / 237(82)

C = 722 / 19434

4.- Result

C = 0.037 cal/g°C

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An increase in aurora activity on Earth could be related to what activity on the Sun?

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Answer:

Explanation:

INCREASE OF SOLAR WINDS When the sun is more active

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3 years ago

A student reduces the temperature of a 300 cm3 balloon from 60°c to 30°

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We can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = 303.15 x 300 / 333.15

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3 years ago

At the most basic level, in order to produce oxygen and a carbohydrate (comprised of carbon, hydrogen, and oxygen), carbon dioxi

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At the most basic level, in order to produce oxygen and a carbohydrate (comprised of carbon, hydrogen, and oxygen), carbon dioxide must react with water as a source of hydrogen.

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Commonly used notation for photosynthesis is:

6CO $_{2}$ + 6H $_{2}$ O C $_{6}$ H $_{12}$ O $_{6}$ + 6O $_{2}$

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2 years ago

Please help me solve this!

yulyashka [42]

Answer : The image is attached below.

Explanation :

For $O_3$ :

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = $\frac{m}{M}=\frac{24g}{48g/mol}=0.5mol$

Number of particles, N = $n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}$

For $NH_3$ :

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = $\frac{m}{M}=\frac{170g}{17g/mol}=10mol$

Number of particles, N = $n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}$

For $F_2$ :

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = $\frac{m}{M}=\frac{38g}{38g/mol}=1mol$

Number of particles, N = $n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}$

For $CO_2$ :

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = $n\times M=0.10mol\times 44g/mol=4.4g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}$

For $NO_2$ :

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = $n\times M=0.20mol\times 46g/mol=9.2g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}$

For $Ne$ :

Molar mass, M = 20 g/mol

Number of particles = $1.5\times 10^{23}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol$

Mass, m = $n\times M=0.25mol\times 20g/mol=5g$

For $N_2O$ :

Molar mass, M = 44 g/mol

Number of particles = $1.2\times 10^{24}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol$

Mass, m = $n\times M=1.9mol\times 44g/mol=83.6g$

For unknown substance:

Number of particles = $3.0\times 10^{23}$

Mass, m = 8.5 g

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol$

Molar mass, M = $\frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol$

The substance is $NH_3$ .

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3 years ago

Calculate the enthalpy change for the melting of a 25.0g ice cube

joja [24]

Enthalpy change = mass * specific heat capacity * temperature change

Mass is 25.0 (given)
Specific heat capacity is 2.108 (checked online)
Temperature change : none
So input these into the equation :

25.0 * 2.108 = 52.7

7 0

3 years ago