<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
Answer:
Explanation:
Aly's teacher said some bacteria can be as small as 0.00365 cm long.
In this problem, we need to write the length of the bacteria written in scientific notation. Any number can be written in the form of scientific notation as follows :
a is a real no and b is an integer
In the given number, there are two zeroes before 3. We need to shift the decimal after 3. It can be done as follows :
Hence, this is the required solution.
Answer:
Insoluble in water:
Explanation:
Water turns out to be a good solvent for ionic substances, or in general, polarized covalent substances. On the other hand, it is not a good solvent for non-polar substances, these being the vast majority of covalent substances.
