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Water has a high specific heat because: Select one: a. it is a poor insulator b. hydrogen bonds must be broken to raise its temp

Crank

<h2>The required "option is b) hydrogen bonds must be broken to raise its temperature.</h2>

Explanation:

Water has high specific heat due to hydrogen bonds present in it.
The Ionisation of water does not affect the specific heat of the water.
On decreasing the temperature, there is the formation of bonds hence option (d) is wrong.
On increasing the temperature, there is the breaking of bonds hence option (b) is correct.

3 0

4 years ago

What would be the volume in millilitres of a blood sample of 2.15 microliters (ul)?

Zanzabum

2.15 x 10⁻³mL

Explanation:

Given parameter:

Volume of blood sample in uL = 2.15uL

Conversion uL → mL

micro- and milli- are both prefixes of sub-units.

liter is a unit of volume of a substance.

micro - is 10⁻⁶

milli- is of the order 10⁻³

The problem is converting from micro to milli:

if we multiply 10⁻⁶ by 10³ we would have our milli;

1000uL = 1mL

2.15uL : 2.15uL x $\frac{1mL}{1000uL}$ = 2.15 x 10⁻³mL

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4 0

3 years ago

1 point

inysia [295]

Answer:

The fungus has grown larger

Explanation:

Because where the orange is in the fridge and even normally you out oranges on the counter or in a bowl, where it's in the fridge it got old faster.

8 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$