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tatyana61 [14]
2 years ago
10

Large masses Of flowing ice called blank are typically found near earths poles and in other cold regions

Chemistry
2 answers:
Leona [35]2 years ago
7 0
The answer is iceberg

Hope this helps :)
bixtya [17]2 years ago
5 0
Larges of flowing ice called glaciers are typically found near Earth's poles and other cold regions.

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What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce m
Rus_ich [418]

Answer:

The percent yield of a reaction is 48.05%.

Explanation:

WO_3+3H_2\rightarrow W+3H_2O

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g

Theoretical yield of water : T

Moles of tungsten(VI) oxide = \frac{51.5 g}{232 g/mol}=0.2220 mol

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

\frac{3}{1}\times 0.2220 mol=0.6660 mol

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%

The percent yield of a reaction is 48.05%.

7 0
3 years ago
Read 2 more answers
Which word best discribes entropy
salantis [7]

Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system.
4 0
3 years ago
1-Propynyllithium reacts with (R,R)-2,3-dimethyloxacyclopropane in a stereoselective reaction. Draw a curved arrow mechanism and
QveST [7]

Answer:

Reference image attached

Explanation:

Please see the attached image.

6 0
2 years ago
A solid powder is known to be a mixture of NaCl and Na2CO3, but the relative amounts of each compound in the sample are unknown.
AveGali [126]

Answer:

Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.

Explanation:

Na_2CO_3(aq) +2 HCl(aq)\rightarrow 2NaCl(aq) + H_2O(l)+CO_2(g)

Molarity of HCl solution = 0.1174 M

Volume of HCl solution = 83.15 mL = 0.08315 L

Moles of HCl = n

molarity=\frac{moles}{Volume (L)}

0.1174 M=\frac{n}{0.08315 L}

n=0.1174 M\times 0.08315 L=0.009762 mol

According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then 0.009762 mol of HCl will recat with:

\frac{1}{2}\times 0.009762 mol=0.004881 mol

Moles of Sodium carbonate = 0.004881 mol

Volume of the sodium carbonate containing solution taken = 1L

Concentration of sodium carbonate in the solution before the addition of HCl:

[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L

4 0
3 years ago
2. When heated above 500 ºC, potassium nitrate decomposes according to the equation below. 4KNO3 2K2O + 2N2 + 5O2A.If oxygen is
AysviL [449]

Answer:

(a) The rate of formation of K2O is 0.12 M/s.

The rate of formation of N2 is also 0.12 M/s

(b) The rate of decomposition of KNO3 is 0.24 M/s

Explanation:

(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.

Rate of formation of O2 is 0.3 M/s

Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s

Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.

Rate of formation of N2 = (2×0.3/5) = 0.12 M/s

(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.

Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s

3 0
2 years ago
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