All categories

3 years ago

Is yeast the same as baking powder

2 answers:

5 0

Answer: "Yeast, baking soda, and baking powder are all leavening agents used in baking. Yeast reacts with sugar, causing it to ferment. The fermentation then results in the production of carbon dioxide. The carbon dioxide gets trapped in bread dough and becomes little air bubbles responsible for making bread rise.

Explanation:

Lunna [17]3 years ago

4 0

Answer: No yeast is not the same thing as baking powder but they are both leavening agents.

You might be interested in

how many grams of water are produced when 15 grams of magnesium hydroxide are reacted with sulfuric acid

Mila [183]

<h3>Answer:</h3>

9.26 g of water

<h3>Explanation:</h3>

We are given;

15 grams of magnesium hydroxide;

Required to calculate the mass of water produced;

We will use the following steps;

<h3>Step 1: Write the balanced equation for the reaction</h3>

Magnesium hydroxide reacts with sulfuric acid to produce water and magnesium sulfate;

Therefore, the equation of the reaction is;

Mg(OH)₂(aq) + H₂SO₄(aq) → MgSO₄(aq) + 2H₂O(l)

<h3>Step 2: Determine the number of moles of Mg(OH)₂ used </h3>

Number of moles = Mass ÷ Molar mass

Molar mass of Mg(OH)₂ =58.320 g/mol

Therefore;

Moles of Mg(OH)₂ = 15 g ÷ 58.320 g/mol

= 0.257 moles

<h3>Step 3: Calculate the moles of water produced</h3>

From the equation 1 mole of Mg(OH)₂ reacts to produce 2 moles of water;

Therefore; Moles of water = moles of Mg(OH)₂ × 2

Moles of water = 0.257 moles × 2

= 0.514 moles

<h3>Step 4: Calculate the mass of water produced;</h3>

Mass = Moles × Molar mass

Molar mass of water = 18.02 g/mol

Therefore;

Mass of water = 0.514 moles × 18.02 g/mol

= 9.26 g

Therefore, 9.26 g of water will be produced

3 0

3 years ago

What is the mass in grams of 250 ml of PH3 at STP?

SIZIF [17.4K]

We can find the number of moles of PH₃ using ideal gas law equation
PV = nRTwhere
P - standard pressure - 101 325 Pa
V - volume - 250 x 10⁻⁶ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 273 K
substituting the values in the equation
101 325 Pa x 250 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 273 K
n = 0.011 mol
therefore mass of PH₃ = 0.011 mol x 34 g/mol = 0.374 g
mass of PH₃ is 0.374 g

8 0

4 years ago

Consider the following reaction:

slava [35]

Answer:

Option D is correct = 58 g

Explanation:

Data Given:

mass of LiOH = 120 g

Mass of Li3N= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)

1 mol 3 mol

Convert moles to mass

Molar mass of LiOH = 24 g/mol

Molar mass of Li₃N = 35 g/mol

So,

Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)

1 mol (35 g/mol) 3 mol (24 g/mol)

35 g 72 g

So if we look at the reaction 35 g of Li₃N react with water and produces 72 g of LiOH , then how many g of Li₃N will be react to Produce by 120 g of LiOH

For this apply unity formula

35 g of Li₃N ≅ 72 g of LiOH

X of Li₃N ≅ 120 g of LiOH

By Doing cross multiplication

Mass of Li₃N = 35 g x 120 g / 72 g

mass of Li₃N = 58 g

120 g of LiOH will produce from 58 g of Li₃N

So,

Option D is correct = 58 g

5 0

3 years ago

What is the relationship between a compound and the elements it is made from? (4 points)

Serggg [28]

Answer:

The compound may have properties that are very different from those of the elements.

Explanation:

For example - common salt, sodium chloride ( NaCl) has very different properties from sodium and chlorine

7 0

3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

4 years ago