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Lostsunrise [7]
3 years ago
14

How many grams of NaBr are in 1.69 moles of NaBr?

Chemistry
1 answer:
MrMuchimi3 years ago
4 0

Answer:

201.671789199999

Explanation:

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C3H8+5O2 --> 3CO2 + 4H2O
Tresset [83]

Answer:

3, 4 ,1

Explanation:

6 0
3 years ago
How may moles are in 88.4 grams of Al(OH)3?
mash [69]

Answer:

1.133mol Al(OH)3

Explanation:

88.4/78.003 = 1.133

5 0
3 years ago
What 3 variables were changed in the pendulum experiment?
Paraphin [41]

Answer:

In this experiment, the period of the pendulum is the dependent variable. There are three independent variables: the pendulum mass, the amplitude of the swing, and the length of the pendulum string.

Explanation:

5 0
3 years ago
What kind of chemical reaction is this
mezya [45]

Answer: from the lies this is synthesis reaction

Explanation:

It's also oxidation-reduction reaction. Li is oxidised and

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7 0
3 years ago
Read 2 more answers
QUESTION 3 Consider a solution containing 0.80 M NaF and 0.80 M HF. Calculate the moles of HF and the concentration of HF after
Lisa [10]

Answer:

0.056moles HF and 0.70M

Explanation:

When a strong acid is added to a buffer, the acid reacts with the conjugate base.

In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:

NaF + HCl → HF + NaCl

Initial moles of NaF and HF in 60.0mL of solution are:

NaF:

0.0600L × (0.80mol / L)= 0.048 moles NaF

HF:

0.0600L × (0.80mol / L)= 0.048 moles HF

Then, the added moles of HCl are:

0.0200L × (0.40mol / L) = 0.008 moles HCl.

Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:

<em>0.056moles HF</em>

<em></em>

In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:

0.056mol HF / 0.0800L = <em>0.70M</em>

6 0
3 years ago
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