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DaniilM [7]
3 years ago
12

An unknown solution has a pH of 2. How would you classify this solution?

Chemistry
1 answer:
devlian [24]3 years ago
7 0
Acidic because anything over a ph of 7 would be basic but anything below 7 would be acid and if it has a ph of 7 it would be neutral....Hope I helped
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is the metal clip on the seatbelt in the summertime have a higher specfic heat or does the fabric on the seat belt?
SOVA2 [1]
Yes because metal is more attracted to heat from the sun
5 0
3 years ago
A. How many hydrogen atoms are on the Reactants side?
patriot [66]
The reactants side are the first ones in a chemical reaction. 
8 0
3 years ago
A 100.0-g sample of sodium hydroxide contains sodium, oxygen, and hydrogen. The
enyata [817]

Answer:

57.47 grams of sodium is in the sample.

Explanation:

Given that a 100.0-g sample of sodium hydroxide contains sodium, oxygen, and hydrogen, and the sample contains 2.53 g of hydrogen and 40.0 g of oxygen, to determine what mass of sodium is in the sample the following calculation must be performed:

100 - (40 + 2.53) = X

100 - 42.53 = X

57.47 = X

Therefore, 57.47 grams of sodium is in the sample.

3 0
2 years ago
Calculate the work in kilojoules done during a synthesis of ammonia in which the volume contracts from 7.4 l to 4.5 l at a const
aksik [14]
<span>The work done by the gas can be derived from this formula: Work done = P dV ( this is the same as PdeltaV) Now remember P = 44atm and Now dV ( deltaV) = 4.5 - 7.4 = -2.9L ( its contracting) = -2.9 x 10 ^-3m^3 Now remember 1 atm = 101300 pascals Therefore work done by gas = P dV = 44 * -2.9 * x 10 ^-3 *101300 Therefore the above result becomes: = -12925.88 = -13000J to 2 significant figures</span>
8 0
3 years ago
Read 2 more answers
Iodine is prepared both in the laboratory and commercially by adding Cl 2 ( g ) to an aqueous solution containing sodium iodide.
Tamiku [17]

Answer:

101.69 grams.

Explanation:

The reaction involved in the preparation of Iodine is mentioned in the question as -

2 NaI_(aq_) +Cl_2_(g_) ⇒ I_2_(s_) +2NaCl_(aq_)

Looking at the stoichiometric coefficients we can say that 2 moles of NaI is required for the preparation of one mole of I_2. In terms of mass we can say that, 299.78 grams of NaI is required for the production of 253.81 grams of I_2.

Now, since 253.81 grams of I_2 is produced by 299.78 grams of NaI

So,

Grams of NaI required for the production of 86.1 grams of I_2 =

\frac{299.78\times86.1}{253.81} =101.69 grams

7 0
3 years ago
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