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1 year ago

In 1-2 sentences, explain why weather can be predicted only as probable, not definite. (2 points) Plaese help me

1 answer:

8 0

Weather can be predicted only as probable, not definite because the daily weather conditions depends on winds and storms.

It is common observation that weather forecasts are often given as a probability and never in definite terms.

This is because, the weather condition at anytime depends on the temperature of the earth's atmosphere which causes air masses to move leading to wind.

The temperature of the earth's atmosphere changes frequently hence weather conditions also change frequently.

Learn more: brainly.com/question/21209813

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How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?

lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

N₂ + 3 H₂ ------—> 2 NH₃

1 mol 3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

N₂ + 3 H₂ ------—> 2 NH₃

1 mol (28 g/mol) 3 mol(2g/mol)

28 g 6 g

28 grams of N₂ react with 6 g of H₂

if 28 grams of N₂ produces 6 g of H₂ so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

28 g of N₂ ≅ 6 g of H₂

X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

X g of N₂ = 28 g x 14.2 g / 6 g

X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

N₂ + 3 H₂ ------—> 2 NH₃

6 g 2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

6 g of H₂ ≅ 2 mol of NH₃

14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

X mol of NH₃= 14.2 g x 2 mol / 6 g

X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of NH₃

Molar mass of NH₃ = 14 + 3(1)

Molar mass of NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

mass in grams = 4.73 mole x 17 g/mol

mass in grams = 80.41 g

mass of NH₃= 80.41 g

3 0

1 year ago

Unit Conversion Help Thank you

AlekseyPX

Answer : 1721.72 g/qt are in 18.2 g/cL

Explanation :

As we are given: 18.2 g/cL

Now we have to convert 18.2 g/cL to g/qt.

Conversions used are:

(1) 1 L = 100 cL

(2) 1 L = 1000 mL

(3) 1 qt = 946 qt

The conversion expression will be:

$\frac{18.2g}{1cL}\times \frac{100cL}{1L}\times \frac{1L}{1000mL}\times \frac{946mL}{1qt}$

$=1721.72\text{ g/qt}$

Therefore, 1721.72 g/qt are in 18.2 g/cL

5 0

2 years ago

II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo

Leto [7]

Answer:

$Excess=3.53g$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg(s)+O_2(g) \rightarrow 2MgO(s)$

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

$n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO$

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

$m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg$

Thus, the mass in excess is:

$Excess=3.86g-0.334g\\\\Excess=3.53g$

Regards!

5 0

2 years ago

What is the yield of uranium from 2.50 kg U3O8?

mr_godi [17]

Answer: Thus the yield of uranium from 2.50 kg $U_3O_8$ is 2.12 kg

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $(6.023\times 10^{23})$ of particles.