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8_murik_8 [283]

3 years ago

14

Many bacteria cannot reproduce and cooler temperatures in our destroyed at high temperatures. How do you humans take advantage o

f this fact when preparing and storing food?

1 answer:

12345 [234]3 years ago

3 0

freezer and refrigerator too cold for most bacteria to grow .oven too hot for bacteria to survive

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The element phosporus is located to the right of the element magnesium along the same period in the periodic table. Which best d

Rus_ich [418]

B. Phosphorus has a larger atomic mass.

3 0

3 years ago

Read 2 more answers

How many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?

zepelin [54]

Answer:

Explanation:

The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.

5 0

2 years ago

As the pendulum moves from point 2 to point 3, what happens to its mechanical energy?potential energy is converted to kinetic en

ahrayia [7]

Answer: Potential energy is converted to kinetic energy and back again.

Explanation:At points 1 and 3, the pendulum stops moving, and its mechanical energy is purely potential. At point 2, the pendulum is moving the fastest, and its mechanical energy is purely kinetic. Therefore, as the pendulum moves from point 1 to point 3, its potential energy is first converted to kinetic energy, then back to potential.

3 0

3 years ago

S(s)+3F2(g)->SF6(g) how many mol of F2 are required to react completely with 2.30 mol of S?

Brilliant_brown [7]

Answer: There are 6.9 mol of $F_{2}$ are required to react completely with 2.30 mol of S.

Explanation:

The given reaction equation is as follows.

$S(s) + 3F_{2}(g) \rightarrow SF_{6}(g)$

Here, 1 mole of S is reaction with 3 moles of $F_{2}$ which means 1 mole of S requires 3 moles of $F_{2}$ .

Therefore, moles of $F_{2}$ required to react completely with 2.30 moles S are calculated as follows.

$1 mol S = 3 mol F_{2}\\2.30 mol S = 3 mol F_{2} \times 2.30 \\= 6.9 mol F_{2}$

Thus, we can conclude that there are 6.9 mol of $F_{2}$ are required to react completely with 2.30 mol of S.

3 0

3 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

3 years ago