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3 years ago

Which stament about energy is mostly correct

1 answer:

6 0

I would say C is the most correct.
In D it depends on what water source you're using. Let's say it is a waterfall, then the source of the water (melting ice or a lake) may disappear in the future.
If you're using underwater "windmills" placed in the ocean, then you would expect it to last a while as the ocean will not disappear in the near future.

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How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago

What substance is added to cordial to make it more dilute?

shusha [124]

The answer is water.

That is water is added to cordial to make it more dilute.

Cordial is basically used to describe sweeter distilled spirits . Or it can be used to describe non-alcoholic drink. These drinks can be diluted using water.

So the answer is water, that is water is added to cordial to make it more dilute.

7 0

3 years ago

What is the pH of a solution of 0.600 M K2HPO4, potassium hydrogen phosphate?

AleksAgata [21]

When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2 
pKa2 = -log(Ka2) ; pKa3 = -log(Ka3) 
so, for this problem, REGARDLESS OF THE CONCENTRATION GIVEN, the answer is: 
pH = (7.2076+12.3767) / 2 
pH = 9.79

5 0

4 years ago

Read 2 more answers

A chemist titrates 90.0 mL of a 0.5870 M acetic acid (HCH, CO) solution with 0.4794M NaOH solution at 25 °C. Calculate the pH at

hjlf

Answer:

9.09

Explanation:

Please kindly check attachment for the step by step solution of the given problem.