Answer:
![\boxed {\boxed {\sf 333 \ grams}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20333%20%5C%20grams%7D%7D)
Explanation:
We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.
![Q= mc \Delta T](https://tex.z-dn.net/?f=Q%3D%20mc%20%5CDelta%20T)
The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.
- ΔT= final temperature - initial temperature
The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.
- ΔT= 88.9 °C - 58.8 °C = 30.1 °C
Now we know three variables:
- Q= 4500.0 J
- c= 0.4494 J/g°C
- ΔT = 30.1 °C
Substitute these values into the formula.
![4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)](https://tex.z-dn.net/?f=4500.0%20%5C%20J%20%3D%20m%20%280.4494%20%5C%20J%2Fg%20%5Ctextdegree%20C%29%2830.1%20%5Ctextdegree%20C%29)
Multiply on the right side of the equation. The units of degrees Celsius cancel.
![4500.0 \ J = m (13.52694 J/g)](https://tex.z-dn.net/?f=4500.0%20%5C%20J%20%3D%20m%20%2813.52694%20J%2Fg%29)
We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g
![\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}](https://tex.z-dn.net/?f=%5Cfrac%20%7B4500.0%20%5C%20J%20%7D%7B13.52694%20J%2Fg%7D%3D%20%5Cfrac%7Bm%20%2813.52694%20J%2Fg%29%7D%7B13.52694%20J%2Fg%7D)
The units of Joules cancel.
![\frac {4500.0 \ J }{13.52694 J/g}= m](https://tex.z-dn.net/?f=%5Cfrac%20%7B4500.0%20%5C%20J%20%7D%7B13.52694%20J%2Fg%7D%3D%20m)
![332.6694729 \ g =m](https://tex.z-dn.net/?f=332.6694729%20%5C%20g%20%3Dm)
The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.
![333 \ g \approx m](https://tex.z-dn.net/?f=333%20%5C%20g%20%5Capprox%20m)
The mass of the sample of metal is approximately <u>333 grams.</u>
Answer: The molarity of the malonic acid solution is 0.08335 M
Explanation:
![H_2C_3H_2O_4 +2NaOH\rightarrow Na_2C_3H_2O_4+2H_2O](https://tex.z-dn.net/?f=H_2C_3H_2O_4%20%2B2NaOH%5Crightarrow%20Na_2C_3H_2O_4%2B2H_2O)
To calculate the molarity of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![H_2C_3H_2O_4](https://tex.z-dn.net/?f=H_2C_3H_2O_4)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
![n_1=2\\M_1=?\\V_1=12.70mL\\n_2=1\\M_2=0.1000M\\V_2=21.17mL](https://tex.z-dn.net/?f=n_1%3D2%5C%5CM_1%3D%3F%5C%5CV_1%3D12.70mL%5C%5Cn_2%3D1%5C%5CM_2%3D0.1000M%5C%5CV_2%3D21.17mL)
Putting values in above equation, we get:
![2\times M_1\times 12.70=1\times 0.1000\times 21.17\\\\M_1=0.08335M](https://tex.z-dn.net/?f=2%5Ctimes%20M_1%5Ctimes%2012.70%3D1%5Ctimes%200.1000%5Ctimes%2021.17%5C%5C%5C%5CM_1%3D0.08335M)
Thus the molarity of the malonic acid solution is 0.08335 M
A physical change.
Physical change = can be reverted + doesn't change substance
Chemical = changes substance
Answer:
According to avogadro's law, 1 mole of every substance contains avogadro's number
of particles and weighs equal to its molecular mass.
To calculate the moles, we use the equation:
a. moles in 14.08 g of
= ![\frac{14.08g}{342.3g/mol}=0.04113moles](https://tex.z-dn.net/?f=%5Cfrac%7B14.08g%7D%7B342.3g%2Fmol%7D%3D0.04113moles)
molecules in 14.08 g of
= ![0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}](https://tex.z-dn.net/?f=0.04113%5Ctimes%206.023%5Ctimes%2010%5E%7B23%7D%3D0.2477%5Ctimes%2010%5E%7B23%7D)
b. moles in 17.75 g of NaCl = ![\frac{17.75g}{58.5g/mol}=0.3034moles](https://tex.z-dn.net/?f=%5Cfrac%7B17.75g%7D%7B58.5g%2Fmol%7D%3D0.3034moles)
molecules in 17.75 g of
= ![0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}](https://tex.z-dn.net/?f=0.3034%5Ctimes%206.023%5Ctimes%2010%5E%7B23%7D%3D1.827%5Ctimes%2010%5E%7B23%7D)
formula units 17.75 g of
= ![0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}](https://tex.z-dn.net/?f=0.3034%5Ctimes%206.023%5Ctimes%2010%5E%7B23%7D%3D1.827%5Ctimes%2010%5E%7B23%7D)
c. moles in 20.06 g of
= ![\frac{20.06g}{249.68g/mol}=0.08034moles](https://tex.z-dn.net/?f=%5Cfrac%7B20.06g%7D%7B249.68g%2Fmol%7D%3D0.08034moles)
formula units in 20.06 g of
= ![0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}](https://tex.z-dn.net/?f=0.08034%5Ctimes%206.023%5Ctimes%2010%5E%7B23%7D%3D0.4839%5Ctimes%2010%5E%7B23%7D)
Answer:
D
Explanation:
We must study the reaction pictured in the question closely before we begin to attempt to answer the question.
Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.
Hence we must show electron movements in both species using a half arrow.