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kkurt [141]
3 years ago
15

An acid

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
4 0
An acid a. has a high pH in solution.
b. turns blue litmus paper to red.
c. releases hydroxyl ions in solution.
d. has more hydroxyl than hydrogen (or hydronium) ions.

<span>An acid </span>turns blue litmus paper to red. The answer is letter B.
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I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.
Alex_Xolod [135]

Answer:

175

Explanation:

3 0
3 years ago
What is the compound name for the formula [Ru(en)2Cl2]2+ and [Co(en)Cl2Br]-
mixer [17]

Answer:2C12

Explanation:because if u divide the the equation by 2 u well get the same answer you can multiply but that will take longer with the invisible zeros and stuff like that but the answer is 2c12 and on the real FSA u can pull out ur phone and cheat just make sure the teacher is not watching just playing with you DONT DO THAT YOU WILL GET IN SERIOUS TROBLE KIDS DO NOT DO THIS KN THE REAL DAY OF THE FSA PLEASE DONT DO IT YOU WILL GET IN SERIOUS TROUBLE WELL THATS IT FOR TODAY SO HAVE A GREAT DAY THIS IS TO PUT PRESSURE ON YOU

3 0
3 years ago
Please help asap thank you.
Aleks [24]

Answer:

a)0.816g/l

b)0.241g/l

8 0
3 years ago
If 0.092J of heat causes a 0.267 degree C temperature change, what mass of water is present?
Thepotemich [5.8K]

Answer:

0.082g

Explanation:

The following data were obtained from the question:

Heat (Q) = 0.092J

Change in temperature (ΔT) = 0.267°C

Specific heat capacity (C) of water = 4.184J/g°C

Mass (M) =..?

Thus, the mass of present can be obtained as follow:

Q = MCΔT

0.092 = M x 4.184 x 0.267

Divide both side by 4.184 x 0.267

M = 0.092 / (4.184 x 0.267)

M = 0.082g

Therefore, mass of water was present is 0.082.

6 0
3 years ago
If a pork roast must absorb
Black_prince [1.1K]
Let us  assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2 
heat absorbed by pork = 0.11 x 2217
                                     = 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
                                              = 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
          = 44 x 20.913
          = 920.172g of CO2 emitted 
7 0
3 years ago
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