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3 years ago

The last one thanks guys this is worht 50 point

Chemistry

1 answer:

vovikov84 [41]3 years ago

6 0

5. adaption

6. neck of giraffe

7. endangered

8. extinct

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A gas occupies a 1.5 L container at 25 degrees Celsius and 2.0 atm. If the gas is transferred to a 3.0 L container at the same t

harkovskaia [24]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

P2 = P1 x V1 / V2

P2 = 2.0 x 1.5 / 3

4 0

3 years ago

Read 2 more answers

Isn't gelatin just edible slime

weqwewe [10]

Answer:

Huh?

Explanation:

am confused as u lol

tx for free points

8 0

2 years ago

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How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?

shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically, molarity = $\frac{no. of moles}{Volume (in L) of solution}$

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

molarity = $\frac{no. of moles}{Volume of solution in liter}$

0.0800 M = $\frac{no. of moles}{0.05 L}$

no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

No. of moles = $\frac{mass in grams}{molar mass}$

mass in grams = $no. of moles \times molar mass of CuSO_{4}.5H_{2}O$

= $1.6 mol \times 249.68 g/mol$

= 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0

3 years ago

Chopping onions makes you cry because when you cut into an onion you break its cells. this releases sulfenic acids, which lead t

ruslelena [56]

Whats the question ?

8 0

3 years ago

Just as one dozen eggs always has 12 eggs in it, one mole of a

zzz [600]

Answer:

$6.022x10^{23}atoms \ Al$

Explanation:

Hello,

In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:

$27.0gAl*\frac{1molAl}{27.0gAl} *\frac{6.022x10^{23}atoms \ Al}{1molAl} \\\\=6.022x10^{23}atoms \ Al$

Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.

Best regards.

8 0

3 years ago