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3 years ago

Data definition from scientific method

Chemistry

1 answer:

mixas84 [53]3 years ago

8 0

Answer:

a method of research in which a problem is identified, relevant data are gathered, a hypothesis is formulated from these data, and the hypothesis is empirically tested.

Hope this helps:)

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Balance the equation by inserting coefficients as needed. equation: C 2 H 6 O+O 2 longrightarrow CO 2 +H 2 O

shepuryov [24]

The coefficients needed to balance the equation would be 2, 7, 4, and 6 respectively.

<h3>Balancing chemical equation</h3>

This requires that the number of atoms of each element is the same both on the reactant and product's sides.

Hence, the balanced equation of the reaction becomes:

$2C_2 H_6 O+ 7O_2 --- > 4CO_2 +6H_2 O$

Thus, the coefficient for each species would be 2, 7, 4, and 6 respectively.

More on balancing chemical equations can be found here: brainly.com/question/15052184

#SPJ1

4 0

3 years ago

An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin

dolphi86 [110]

Answer:

The theoretical yield of urea = <u>120.35kg</u>

The percent yield for the reaction = <u>72.70%</u>

Explanation:

Lets calculate -

The given reaction is -

$2NH_3(aq)+CO_2$ → $CH_4N_2O(aq)+H_2O (l)$

Molar mass of urea $CH_4N_2O$ = 60g/mole

Moles of $NH_3$ = $\frac{62.8kg/mole}{17g/mole}$ (since $Moles=\frac{mass of substance}{mass of one mole}$ )

= 4011.76 moles

Moles of $CO_2$ = $\frac{105kg}{44g/mole}$

= $\frac{105000g}{44g/mole}$

= 2386.36 moles

Theoritically , moles of $NH_3$ required = double the moles of $CO_2$

but , $4011.76$ , the limiting reagent is $NH_3$

Theoritical moles of urea obtained = $\frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3$

= $2005.88mole CH_4N_2O$

Mass of 2005.88 mole of $CH_4N_2O$ = $2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}$

= 120352.8g

$120352.8g\times \frac{1kg}{1000g}$

= 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = $\frac{87.5kg}{120.35kg}\times100$

72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0

3 years ago

12. How many grams of glucose (C6H1206) is produced if 17.3 mol of H20 is reacted according to this

iren2701 [21]

Answer:

1. 17.3 MOLES OF WATER WILL PRODUCE 518.4 G OF GLUCOSE.

2. 87.4 G OF HNO3 WILL PRODUCE 7.86 G OF NITROGLYCERIN WHEN REACTED WITH EXCESS GLYCEROL.

Explanation:

1. How many grams of glucose is produced from 17.3 mole of water?

Equation:

6CO2 + 6H20 ------> C6H12O6 + 6O2

From the reaction, 6 moles of carbon dioxide reacts with 6 moles of water to produce 1 mole of glucose

So therefore,

6 moles of water = 1 mole of glucose

Since 17.3 mole of water was used, we can first calculate the number of moles of glucose produced:

Then, we have:

6 moles of water = 1 mole of glucose

17.3 moles of water = ( 17.3 * 1/ 6) moles of glucose

= 2.883 moles of glucose

So we say 17.3 moles of water produces 2.883 moles of glucose

At standard conditions, 1 mole of a substance is its molar mass

Molar mass of water = 18 g/mol

Molar mass of glucose = 180 g/mol

From the reaction:

17.3 moles of water produces 2.883 moles of glucose

17.3 * 18 g of water produces 2.833 * 180 g of glucose

= 518.94 g of glucose.

From 17.3 mole of water, 518.4 g of glucose will be produced.

C3H5(OH)3 + 3HNO3 -------> C3H5(ONO2)3 + 3H20

3 moles of HNO3 reacts to produce 1 mole of nitroglycerin

Molar mass of HNO3 = ( 1 + 14 + 16*3) = 63 g/mol

Molar mass of nitroglycerin = ( 12 *3 + 1 *5 + 16*6 + 14*3) = 179 g/mol

3 moles of HNO3 = 1 mole of nitroglycerin

3 * 63 g of HNO3 = 179 g of nitroglycerin

if 87.4 g of HNO3 were to be reacted, we have:

189 g of HNO3 = 179 g of nitroglycerin

87.4 g of HNO3 = ( 87.4 * 179 / 189) of nitroglycerin

= 7.86 g of nitroglycerin

So therefore, 87.4 g of HNO3 will produce 7.86 g of nitroglycerin when reacted with excess glycerol.

4 0

4 years ago

Which would melt first, the element

balandron [24]

Explanation:

First, get both values into the same units

To covert from Kelvin to degrees C, subtract 273.15

1210 - 273.15 =936.85

Melting point of germanium=936.85 degrees C

Melting point of gold= 1064 degrees C

Therefore germanium would melt first because it has the lower melting point.

4 0

3 years ago

Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp

weeeeeb [17]

Answer : The grams of $N_2$ consumed is, 89.6 grams.

Solution : Given,

Mass of $CaCN_2$ = 265 g

Molar mass of $CaCN_2$ = 80 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate the moles of $CaCN_2$ .

$\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles$

The given balanced reaction is,

$CaC_2+N_2\rightarrow CaCN_2+C$

from the reaction, we conclude that

As, 1 mole of $CaCN_2$ produces from 1 mole of $N_2$

So, 3.2 moles of $CaCN_2$ produces from 3.2 moles of $N_2$

Now we have to calculate the mass of $N_2$

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

$\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g$

Therefore, the grams of $N_2$ consumed is, 89.6 grams.

5 0

3 years ago

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