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laila [671]
3 years ago
11

Warm air lifted over a moving cold air mass will produce a front

Chemistry
2 answers:
valina [46]3 years ago
7 0

Answer:

Warm air lifted over a moving cold air mass will produce a cold front

crimeas [40]3 years ago
4 0

Answer:

cold

Warm air lifted over a moving cold air mass will produce a _____ front.

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What is the empirical formula
Radda [10]

Answer:

a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

Explanation:

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2 years ago
Describe how the size of sediment particles affects their movement during deflation.
Oksi-84 [34.3K]
The larger the piece the longer it will take to break down. This is because it has more mass that needs to be broken down.
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3 years ago
Consider the cell notation Cd(s) | Cd(NO3)2(aq) || AgNO3(aq) | Ag(s) with standard reduction potentials for Cd2+ and Ag+ as -0.4
MakcuM [25]

The value of the Gibbs free energy shows us that the reaction is spontaneous.

<h3>What is the Gibbs free energy?</h3>

The Gibbs free energy is a quantity that helps us to be able to determine the spontaneity of a reaction.

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Now;

△G = -nFEcell

△G = -(2 * 96500 * 1.201)

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Learn more about free energy:brainly.com/question/15319033?

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8 0
1 year ago
When water was added to a 4.00 gram mixture of potassium oxalate hydrate (molar mass 184.24 g/mol) and calcium hydrate shown bel
Sliva [168]

Answer:

% (COOK)2H2O = 37.826 %

Explanation:

mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O

∴ mass mix = 4.00 g

∴ mass (CaC2O4)H2O = 1.20 g

∴ Mw (COOK)2H2O = 184.24 g/mol

∴ Mw (CaC2O4)H2O = 146.12 g/mol

∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1

  • % (COOK)2H2O = (mass (COOK)2H2O / mass Mix) × 100

⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)

⇒ mass (COOK)2H2O = 1.513 g

⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100

⇒ % (COOK)2H2O = 37.826 %

7 0
3 years ago
a series circuit contains a generator, two devices, and connecting wires. The resistances of the two devices are 15 ohms and 10
Naddik [55]
Maybe you can apply the Ohms law to this question:

Ohms law is voltage=current (I) x resistance (R)

I'm afraid that is the only concept i know abt electricity. i hope u find a solution;)
7 0
3 years ago
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