Which element exhibits a crystalline structure at STP? 1)F 2)CI 3)Br 4) I

How does a filter separate mixtures like sand and water?

Talja [164]

Answer:

Filtration is a method for separating an insoluble solid from a liquid. When a mixture of sand and water is filtered: the sand stays behind in the filter paper (it becomes the residue ) the water passes through the filter paper (it becomes the filtrate )

Explanation:

8 0

3 years ago

Why are summer days longer than winter days on Earth?

zimovet [89]

Answer:

A. During the summer, Earth's rotational axis is parallel to the Sun's rotational axis.

Explanation:

Actually, though, the Earth is tilted 23.4 degrees! (A circle is 360 degrees.) This tilt is the reason that days are longer in the summer and shorter in the winter. The hemisphere that's tilted closest to the Sun has the longest, brightest days because it gets more direct light from the Sun's rays.

Plz mark me brainliest if correct :)

8 0

3 years ago

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Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.

Jlenok [28]

Answer: The density of gold in $g/cm^3$ is $19.22g/cm^3$

Explanation:

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

$\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}$

We are given:

Density of gold = $1200lb/ft^3$

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into $g/cm^3$ , we get:

$\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3$

Hence, the density of gold in $g/cm^3$ is $19.22g/cm^3$

6 0

3 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago

If 4.0 mol of NO and 4.0 mol of O2 are combined, how many moles

Masja [62]

4.0

i think it has something to do with molar ratios and finding the limiting reactant

4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2

4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2

so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO

once the limiting reactant is found, we can use that data for that substance to calculate the amount of product

4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2

4 0

3 years ago