Answer:
0.3267 M
Explanation:
To solve this problem, first we calculate how many moles of Mn(ClO₄)₂ are contained in 23.640 g of Mn(ClO₄)₂·6H₂O.
Keep in mind that the crystals of Mn(ClO₄)₂ are hydrated, and <em>we need to consider those six water molecules when calculating the molar mass of the crystals</em>.
Molar mass of Mn(ClO₄)₂·6H₂O = 54.94 + (35.45+16*4)*2 + 6*18 = 361.84 g/mol
Now we <u>proceed to calculate</u>:
- 23.640 g Mn(ClO₄)₂·6H₂O ÷ 361.84 g/mol = 0.0653 mol Mn(ClO₄)₂·6H₂O = mol Mn(ClO₄)₂
Now we divide the moles by the volume, to <u>calculate molarity</u>:
- 200 mL⇒ 200/1000 = 0.200 L
- 0.0653 mol Mn(ClO₄)₂ / 0.200 L = 0.3267 M
Answer : Option A) 4.0 X 101
Explanation : As per the measurement chart the greatest metric system is measured in meters.
So the rest values are given in the form of metric values which are obviously less than that of meters. values of centimeters, decimeter and millimeters are less than that of meters.
Voltmeter is the device that is used to measure the potential difference across the battery.
<h2>What are the usage of voltmeter?</h2><h3 /><h3>Usage of Voltmeter</h3>
Voltmeter is an instrument that measures voltages of both direct and alternating electric current. On a scale of voltmeter usually graduated in volts, millivolts (0.001 volt), or kilovolts (1,000 volts).
Voltmeter is connected in parallel form. It has a high resistance so that it takes negligible current from the circuit so we can conclude that Voltmeter is the device that is used to measure the potential difference across the battery.
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y,
for the following reason:
If you look at the combining masses of X and Y in
each of the two compounds,
The first compound contains 0.25g of X combined with
0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3
The second compound contains 0.33 g of X combined with
0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2
Now, you suppose to prepare each of these two
compounds, starting with the same fixed mass of element Y ( I will choose 12g
of Y for an easy calculation!)
The first compound will then contain 4g of X and 12g
of Y
The second compound will then contain 6g of X and
12g of Y
<span>The ratio which combined
the masses of X and the fixed mass (12g) of Y
= 4 : 6
<span>or 2 : 3 </span>
So, the ratio of MOLES of X which combined with the
fixed amount of Y in the two compounds is also = 2 : 3 </span>
The two compounds given with the plausible formula must therefore contain
the same ratio.
