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Vedmedyk [2.9K]
3 years ago
9

How long do fresh cranberries last in the refrigerator?

Chemistry
1 answer:
vodka [1.7K]3 years ago
7 0

Answer:

Three to four weeks

Explanation: I had cranberries and In like four weeks my mom forced me to get rid of them. They do stink when rotten ;(

You might be interested in
What volume of beaker contains exactly 2.23x10^-2 mol of nitrogen gas at STP?
nikklg [1K]

Answer:

V = 0.5 L

Explanation:

Given data:

Moles of nitrogen = 2.23×10⁻² mol (0.0223 mol)

Temperature = 273 K

Pressure = 1 atm

Volume = ?

Solution:

PV = nRT

V = nRT / P

V = 0.0223 mol × 0.0821 atm. mol⁻¹. L . k⁻¹ × 273 K / 1 atm

V = 0.5 L

4 0
4 years ago
1
Levart [38]

Answer:

A

Explanation:

Friction is a force that opposes motion.

5 0
3 years ago
How do you INCREASE potential energy?
fgiga [73]

You have to move higher; potential energy depends on height and mass.

4 0
3 years ago
An aqueous solution contains 0.23 M potassium hypochlorite.
Arturiano [62]

Answer:

0.22 mol HClO, 0.11mol HBr.

0.25mol NH₄Cl, 0.12 mol HCl

Explanation:

A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.

Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO  </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.

Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em>  will produce NH₄⁺. 0.25mol HCl consume all NH₃.

5 0
3 years ago
How many moles of K+ and PO4^3- ions are present in 20.0 mL of a 0.015 M solution of potassium phosphate?
Romashka-Z-Leto [24]

Answer:

n_{K^+}=0.0009molK^+

Explanation:

Hello,

In this case, the first step is to compute the number of moles of potassium phosphate in 20.0 mL (0.020L) of the 0.015-M (mol/L) solution as shown below:

n=0.020L*0.015\frac{mol}{L}=0.0003mol

Thus, these moles correspond to potassium phosphate moles, which molecular formula is K₃PO₄, therefore, one mole of this compound contains three moles of potassium ions as it has three as its subscript in the formula. Thereby, the moles of potassium ions result in:

n_{K^+}=0.0003molK_3PO_4*\frac{3molK^+}{1molK_3PO_4} \\\\n_{K^+}=0.0009molK^+

Best regards.

7 0
4 years ago
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