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11 months ago

8

Any idea what this is You may need to look at the image at varying angles by adjusting your screen.There are 4 words hidden in t

he image.

2 answers:

Yakvenalex [24]11 months ago

6 0

What the person said above me

yan [13]11 months ago

5 0

Answer: Oxygen, Beryllium, Lithium, & Fluorine

Explanation:

theoretically it might not be the answer be

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in this trial, 0.400 M NaOH was added to 40.00 ml of 0.400 M HCl. How many ml of base must be added to cause the colour to chang

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The amount of Ml that is needed

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2 years ago

The electron dot structure for CI is

finlep [7]

The correct option is C.
A Lewis dot diagram is a representation of the valence electron of an atom, which uses dot around the symbol of the atom. Chlorine has seven electrons in its outermost shell, these seven electrons are arranged in form of dot around the atom of chlorine. If you count the number of dot given in option C, you will notice that they are seven.

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3 years ago

Read 2 more answers

Which skin layer holds layers together

lions [1.4K]

The <span>epithelial hold the skin together.</span>

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3 years ago

When the substance dissolves the solution is not transparent true or false <br>

bogdanovich [222]

Answer:

Some substances dissolve when you mix them with water. When a substance dissolves, it might look like it has disappeared, but in fact it has just mixed with the water to make a transparent (see-through) liquid called a solution.

Explanation:

8 0

2 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago