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3 years ago

3. What methods are you using to test this hypothesis? Outline the steps of the procedure in full sentences.

Chemistry

1 answer:

Ulleksa [173]3 years ago

4 0

Answer:

AHYPOTHISES IS A GUESS

Explanation:

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kompoz [17]

Answer:

The first option----A

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3 years ago

Correction?

rjkz [21]

Your answer is correct.

MgO(s) + H2O(l) ----> Mg(OH)2(aq)

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3 years ago

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What volume in mt, of 0.5a M1HCI solution is needed to neutralize 77 ml of 1.54 M NaOH solution?

Rainbow [258]

Answer:

237.2 mL.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HCl = (XMV) NaOH.

For HCl; X = 1, M = 0.5 M, V = ??? mL.

For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.

∴ V of HCl = (XMV) NaOH / (XV) HCl = (1)(1.54 M)(77.0 mL) / (1)(0.5 M) = 237.2 mL.

8 0

3 years ago

1: At which temperature would a reaction withΔH = -102 kJ/mol, ΔS = -0.188 kJ/(mol×K) be spontaneous? 2: At which temperature wo

Naddik [55]

Answer:

1: At temperatures below 542.55 K

2: At temperatures above 660 K

Explanation:

Hello there!

In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:

$\Delta G=\Delta H-T\Delta S$

Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:

$0=\Delta H-T\Delta S\\\\T=\frac{-102kJ/mol}{-0.188kJ/mol-K} \\\\T=542.55K$

It means that at temperatures lower than 542.55 K the reaction will be spontaneous.

$0=\Delta H-T\Delta S\\\\T=\frac{132kJ/mol}{0.200kJ/mol-K} \\\\T=660K$

It means that at temperatures higher than 660 K the reaction will be spontaneous.

Best regards!

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2 years ago

What comes in contact with Sodium Hydroxide to form soap and glycerine it is a chemical change because you cannot reverse this c

svetoff [14.1K]

Answer:

Your answer is triglycerides. Hope this helps.

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3 years ago