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Troyanec [42]
3 years ago
6

PLEASE HELP!!!!

Chemistry
1 answer:
coldgirl [10]3 years ago
7 0

Answer:

1.2 liters.

Explanation:

Focus on the 4th digit: that's the ones column. The 3rd digit is the decimal place, just be sure to round up.

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weqwewe [10]
The finagling in the hole
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3 years ago
Determine the empirical formula of a
k0ka [10]

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold  = 2.6444g

Mass of Chlorine  = 0.476g

Unknown:

Empirical formula  = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements                                     Au                                             Cl

Mass                                         2.6444                                     0.476

Molar mass                                 197                                          35.5

Number of moles                  2.6444/197                                 0.476/35.5

                                                 0.013                                           0.013

Divide by the

smallest                                 0.013/0.013                                 0.013/0.013

                                                       1                                                   1

The empirical formula of the compound is AuCl

3 0
2 years ago
Bob measured out 1.60 grams of sodium. He calculates that 1.60 g of
saul85 [17]

Answer:

84.8%

Explanation:

Step 1: Given data

Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.

Na + 1/2 Cl₂ ⇒ NaCl

According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.

Step 2: Calculate the percent yield.

We will use the following expression.

%yield = real yield / theoretical yield × 100%

%yield = 3.45 g / 4.07 g × 100% = 84.8%

6 0
3 years ago
Complete this equation for the dissociation of NH4NO3(aq).
Delvig [45]
Dissociation
NH₄NO₃(aq) = NH₄⁺(aq) + NO₃⁻(aq)


the hydrolysis of the cation
NH₄⁺(aq) + H₂O(l) = NH₃(aq) + H₃O⁺(aq)
pH<span><7</span>

5 0
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