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Sever21 [200]
3 years ago
9

What could you do to make yeast dough rise more slowly?

Chemistry
2 answers:
Nitella [24]3 years ago
6 0

Answer:

So, if you want to slow down rise without much testing, controlling temperature—allow for a slow rise in the refrigerator or add cold liquid to the dough instead of the usually recommended warm liquid—is the more surefire method. Therefore, your answer should be <u><em>D</em></u>.

Explanation:

olganol [36]3 years ago
3 0

Answer:

d. Reduce the temperature

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PLEASE HURRY THIS IS TIMED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
AURORKA [14]

Answer:

if i remember correctly i beleive its A  1.8 x 10^24

but im not for sure  also i think you forgot the 24

Explanation:

5 0
3 years ago
Classify each salt as acidic, basic, or neutral? -nh4cl -ca(no32
vichka [17]
The pH of salt depends on the component acid and base that comprise them. For example, if the salt is made up of strong acid and weak base then, the salt is acidic. If the salt is formed from strong base and weak acid then, the salt is basic. For this item, NH4Cl is acidic and also Ca(NO3)2 is acidic. 
3 0
2 years ago
Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13
Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\

  =\frac{1571}{10}\\\\=157.1

Calculating the standardn:

\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\

Please find the correct equation in the attached file.

=28.195

For point a:

=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\

For point b:

=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M

For point c:

= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\  M

It is calculated by using the slope value that is 1.75 \times 10^9 M^{-1}. The slope value 1.75 \times 10^9 M^{-1}is ambiguous.

7 0
2 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
What is the main reason scientists perform experiments
Anuta_ua [19.1K]
B, they then interpret that data to find their answers
3 0
3 years ago
Read 2 more answers
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