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Scorpion4ik [409]
3 years ago
14

Examples s name of thosse food items we can store for a month?​

Chemistry
1 answer:
choli [55]3 years ago
3 0

Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders

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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
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Give me some names of a main sequence star other than the sun
ivolga24 [154]
Arietis, camelopardalis, lyncis, scorpii, cancri, canis minoris
7 0
3 years ago
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2N2H4+ N2O4———3N2+4H2O SalmaKhan99 avatar How many grams of N2 gas will be formed by reacting 100g of N2H4 and 200g of N2 Kindly
Alona [7]

Answer:

131.26 g

Explanation:

From the balanced equation,

2 moles of N₂H₄reacts with 1 mole of N₂O₄ to give 3 moles of N₂

Now number of moles of N₂H₄ present in 100 g N₂H₄ is n = 100 g/molar mass N₂H₄.

Molar mass N₂H₄ = 2 × 14.01 g/mol + 1 × 4 g/mol = 28.02 g/mol + 4 g/mol = 32.02 g/mol

n₁ = 100/32.02 = 3.123 mol

Also

Now number of moles of N₂O₄ present in 200 g N₂O₄ is n = 200 g/molar mass N₂O₄.

Molar mass N₂O₄ = 2 × 14.01 g/mol + 16 × 4 g/mol = 28.02 g/mol + 64 g/mol = 92.02 g/mol

n₂ = 200/92.02 = 2.173 mol

Since the mole ratio of N₂H₄  to N₂O₄ is 2 : 1, We require 2 × 2.173 mol N₂H₄  to react with 2.173 mole N₂O₄  

Number of moles of N₂H₄ required is 4.346. But the number of moles of N₂H₄  present is 3.123 so N₂H₄  is the limiting reagent.

So, from the equation, 2 moles of N₂H₄ produces 3 moles of N₂

Therefore number of mole N₂ = 3/2 moles of N₂H₄ = 3/2 × 3.123 mol = 4.6845 mol

From n = m/M where n = number of moles of nitrogen gas = 4.6845 mol and M = molar mass of nitrogen gas = 28.02 g/mol and m = mass of nitogen gas.

m = nM = 4.6845 mol × 28.02 g/mol = 131.26 g

So the mas of nitrogen gas produced is 131.26 g

4 0
4 years ago
Convert 321 millimeters to picometers
pantera1 [17]
3.21e+11 is your answer for 321 millimeters in picometers. Hope this helps!

3 0
3 years ago
Fill in the blanks for the following statements: The rms speed of the molecules in a sample of H2 gas at 300 K will be _________
Anna007 [38]

Answer : The rms speed of the molecules in a sample of H_2 gas at 300 K will be four times larger than the rms speed of O_2 molecules at the same temperature, and the ratio \mu _{rms}(H_2)/\mu _{rms}(O_2) constant with increasing temperature.

Explanation :

Formula used for root mean square speed :

\mu _{rms}=\sqrt{\frac{3RT}{M}}

where,

\mu _{rms} = rms speed of the molecule

R = gas constant

T = temperature

M = molar mass of the gas

At constant temperature, the formula becomes,

\mu _{rms}=\sqrt{\frac{1}{M}}

And the formula for two gases will be,

\frac{\mu _{H_2}}{\mu _{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}

Molar mass of O_2 = 32 g/mole

Molar mass of H_2 = 2 g/mole

Now put all the given values in the above formula, we get

\frac{\mu _{H_2}}{\mu _{O_2}}=\sqrt{\frac{32g/mole}{M_{2g/mole}}}=4

Therefore, the rms speed of the molecules in a sample of H_2 gas at 300 K will be four times larger than the rms speed of O_2 molecules at the same temperature.

And the ratio \mu _{rms}(H_2)/\mu _{rms}(O_2) constant with increasing temperature because rms speed depends only on the molar mass of the gases at same temperature.

5 0
3 years ago
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