Ask question

Ask question

All categories

Citrus2011 [14]

2 years ago

10

Bread is made by mixing flour, water, and yeast together and then baking. How does the example of bread tell the difference betw

een rocks and minerals?
The bread would be a mineral. The flour, water, and yeast would be rocks.

The bread, flour, water, and yeast would all be rocks.

The bread would be a rock. The flour, water, and yeast would be minerals.

The bread, flour, water, and yeast would all be minerals.

1 answer:

Shtirlitz [24]2 years ago

7 0

Answer:

5 measuring spoons

Explanation:

got 100% on edge

You might be interested in

What is the final step in the scientific method

kherson [118]

The final step in the scientific method is the conclusion. The conclusion will either clearly support the hypothesis or it will not. If the results support the hypothesis a conclusion can be written

8 0

3 years ago

Read 2 more answers

Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are

emmasim [6.3K]

When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant.

Let's do the ICE analysis

2 NH₃ ⇄ N₂ + 3 H₂
I 0 1.3 1.65
C +2x -x -3x
-------------------------------------
E 0.1 ? ?

The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 mol
Equilibrium N₂ = 1..3 - 0.05 = 1.25 mol

For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is

$K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}$

Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.

8 0

3 years ago

Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO3 solution hav

Alona [7]

Answer:

pH of HNO₃ having an hydrogen ion concentration of 0.71M is 0.149

Explanation:

HNO₃ (aqueous) ⇄ H⁺ + NO3⁻

The pH is defined as the negative log of the hydrogen ion concentration

pH = - log [H⁺]

From the question, the hydrogen ion concentration is given as 0.71M, therefore

pH = -log [0.71]

= 0.149

3 0

3 years ago

Read 2 more answers

How many Sodium atoms are in 2.00 mols of Na2O

Anna71 [15]

There’s 1024 atoms of sodium in 2 moles of sodium

7 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago