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3 years ago

What is the name of CF7

Chemistry

2 answers:

Varvara68 [4.7K]3 years ago

7 0

Here is the answer in the picture

Wittaler [7]3 years ago

4 0

Answer:

Formula mass in daltons of the chemical component.

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Balance the chemical equation. Based on the equation, how many grams of bromine are produced by the complete reaction of 11 gram

scoundrel [369]

Answer:

Explanation:

The balanced chemical equation is Cl2 + 2KBr → 2KCl + Br2. The amount of bromine is calculated as follows:

11.0 g KBr (1mol KBr/119.002g KBr * 1mol Br2/2mol KBr * 159.808g Br2/1mol Br2= 7.39 g Br2.)

5 0

3 years ago

What happened to the energy of the system when the baking soda mixed with the vinegar?

andrey2020 [161]

Answer:

Explanation:

I hope it help

It help you mixed up

6 0

3 years ago

A solution is made containing 4.6 g of sodium chloride per 250g of water.

lozanna [386]

Answer:

$\large \boxed{1.81 \, \%}$

Explanation:

$\text{Mass percent} = \dfrac{\text{mass of solute} }{\text{mass of solution}} \times 100 \, \%$

Data:

Mass of NaCl = 4.6 g

Mass of water = 250 g

Calculations:

Mass of solution = mass of NaCl + mass of water = 4.6 g + 250 g = 254.6 g.

$\text{\% m/m} = \dfrac{\text{4.6 g} }{\text{254.6 g}} \times 100 \, \% = \mathbf{1.81 \, \%}\\\\\text{The percent by mass of NaCl is $\large \boxed{\mathbf{1.81 \, \%}}$}$

3 0

4 years ago

A student obtains a 10.0g sample of a white powder labeled as BaCl2. After completely dissolving the powder in 50.0mL of distill

LekaFEV [45]

Answer:

Whether barium chloride solution was pure

Explanation:

We may answer whether barium chloride was pure. The sequence of this experiment might be depicted by the following balanced chemical equations:

$BaCl_2 (aq)\rightarrow Ba^{2+} (aq) + 2 Cl^- (aq)$

$Ba^{2+} (aq) + SO_4^{2-} (aq)\rightarrow BaSO_4 (s)$

Having a total sample of 10.0 grams, we would firstly find the mass percentage of barium in barium chloride:

$\omega_{Ba^{2+}} = \frac{M_{Ba}}{M_{BaCl_2}} = \frac{137.327 g/mol}{208.23 g/mol\cdot 100\% = 65.95 \%$

This means in 10.0 g, we have a total of:

$m_{Ba^{2+}} = 10.0 g\cdot 0.6595 = 6.595 g$ of barium cations.

The precipitate is then formed and we measure its mass. Having its mass determined, we'll firstly find the percentage of barium in barium sulfate using the same approach:

$\omega_{Ba} = \frac{137.327 g/mol}{233.38 g/mol}\cdot 100\% = 58.84 \%$

Multiplying the mass we obtained by the fraction of barium will yield mass of barium in barium sulfate. Then:

if this number is equal to 6.595 g, we have a pure sample of barium chloride;
if this number is lower than 6.595 g, this means we have an impure sample of barium chloride, as we were only able to precipitate a fraction of 6.595 g.

3 0

3 years ago

Please someone answer this for me ASAP

Alex777 [14]

Answer:

I'm not 100% sure but I think it is b I'm not great with math tho

6 0

3 years ago