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3 years ago

What is an example of tipping point?

2 answers:

6 0

Answer:Examples of tipping points are everywhere: a glass falling off a table upon tilting; a bacterial population hitting a level where it pushes your body into fever; the boiling point of water, or a cube of ice being thrown into warm water, where it rapidly melts.

Explanation:

sladkih [1.3K]3 years ago

3 0

Answer:

A IPCC AR5 climate tipping point is a critical stage at which global or regional climate shows a major change. 

Explanation:

The atmosphere framework expresses that the exact degrees of environmental change adequate to trigger a tipping point stay unsure, yet that the hazard related to intersection different tipping focuses increments with rising temperature.

Examples are melting of Arctic sea, Collapse of coral leaf due to global warming, eutrophication of the lake, Forest loss, lack of rain, Ocean warming, changes in the temperature of the Ocean current etc.

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A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of

Goryan [66]

Answer:

C₂ H₄ O

Explanation:

1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)

atomic mass of C: 12.0107 g/mol

molar mass of CO₂: 44.01 g/mol

Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂

Solve for x:

x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C

2) Mass of hydrogen (H) in 0.930 g of water (H₂O)

atomic mass of H: 1.00784 g/mol

molar mass of H₂O: 18.01528 g/mol

proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O

Solve for x:

x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H

3) Mass of oxygen (O) in 1.0857 g of pure sample

Mass of O = mass of pure sample - mass of C - mass of H

Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

4) Mole calculations

Divide the mass in grams of each element by its atomic mass:

C: 0.59767 g / 12.0107 g/mol = 0.04976 mol

H: 0.10406 g / 1.00784 g/mol = 0.10325 mol

O: 0.3840 g / 15.999 g/mol = 0.02400 mol

5) Divide every amount by the smallest value (to find the mole ratios)

C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2

H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4

O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

C₂ H₄ O ← answer

3 0

3 years ago

Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some resistant bacteria. The mass

Arisa [49]

Answer:

The turnover number is the maximum substrate quantitiy converted to product per enzyme and per second. It can be calculate as follows:

$k_c = \frac{V_{max}}{[E]}$ with $[E]$ active enzyme concentration.

In this case we have Vmax an data to calculate $[E]$

$moles of E=\frac{10^{-9}g}{29600 g/mol}=3.38*10^{-14}$

$[E]=\frac{3.38*10^{-14}}{0.01 L}=3.38*10^{-12} M$

Now $k_c=\frac{6.8*10^-10}{3.38*10^{-12}}=201.28 min^{-1}$

it is not like any option.

If we assume that $V_{max}$ have the non usual units of $mol L^{-1} min^{-1}$ and it is $6.8*10^-10 mol min^{-1}$

So we need divide by the moles of E (in place of [E])

Now $k_c=\frac{6.8*10^-10}{3.38*10^{-14}}=20128 min^{-1} = 335 s{-1}$

(pass from $min^{-1}$ to $s^{-1}$ dividing by 60)

4 0

3 years ago

Identify the effect of increasing acidity on the solubility of the given compounds Sr(OH)2 S r( O H) Subscript2 ZnS Z n S PbBr2

Dafna1 [17]

Answer:

Solubilities of $Sr(OH)_{2},ZnS,Ca_{3}(PO_{4})_{2} and BaCO_{3}$ increases with increasing acidity

Explanation:

$Sr(OH)_{2}\rightleftharpoons Sr^{2+}+2 OH^{-}$

Concentration of $OH^{-}$ decreases with increasing acidity. Therefore solubility of $Sr(OH)_{2}$ increases to keep constant solubility product.

$ZnS\rightleftharpoons Zn^{2+}+S^{2-}$

Concentration of $S^{2-}$ decreases with increasing acidity due to formation of $H_{2}S$ . therefore solubility of ZnS increases to keep constant solubility product

$PbBr_{2}\rightleftharpoons Pb^{2+}+2Br^{-}$

Increasing acidity have no effect on this above equilibrium. Therefore solubility of $PbBr_{2}$ remains constant.

$Ca_{3}(PO_{4})_{2}\rightleftharpoons 3Ca^{2+}+2PO_{4}^{3-}$

Concentration of $PO_{4}^{3-}$ decreases with increasing acidity due to formation of $H_{3}PO_{4}$ . Therefore solubility of $Ca_{3}(PO_{4})_{2}$ increases to keep constant solubility product.

$NaI\rightleftharpoons Na^{+}+I^{-}$

Increasing acidity have no effect on this above equilibrium. Therefore solubility of NaI remains constant.

$BaCO_{3}\rightleftharpoons Ba^{2+}+CO_{3}^{2-}$

Concentration of $CO_{3}^{2-}$ decreases with increasing acidity due to formation of $H_{2}CO_{3}$ . Therefore solubility of $BaCO_{3}$ increases to keep constant solubility product.

5 0

3 years ago

According to legend, archimedes, who was a famous greek mathematician, was given a crown, which was supposed to be made of pure

aniked [119]

Determining the volume of water displaced by the crown and using the gold density to determine the mass of the crown equals the mass got from weighing the crown on a weighing machine. if the mass matches the mass got from the weighing machine then the crown is made of pure gold if otherwise, the gold is not pure.

3 0

4 years ago

Read 2 more answers

The positvie part of the atom found in the center of the atom

vladimir2022 [97]

That would be protons ! They are positive

6 0

2 years ago

Read 2 more answers