The heat of fusion for water is 80. cal/g. how many calories of heat are released

In the formate ion, hco2− which of the atoms in the ion have pπ orbitals?

kirza4 [7]

I think it is halico acid

5 0

3 years ago

What is a potential benefit of fuel cell cars?

77julia77 [94]

Answer:

no recharging necessary

3 0

3 years ago

Read 2 more answers

Limit sodium intake to less than ____________ per day. Limit saturated fat intake to less than ____________ of daily kilocalorie

kotykmax [81]

Answer:

Limit sodium intake to less than 2,300mg per day. Limit saturated fat intake to less than 10% of daily kilocalories. Limit intake of added sugars to less than 5% of daily kilocalories. Limit intake of alcohol to less than 3 drinks per day for men or less than 2 drinks per day for women.

Explanation:

The human body has been designed such that there are dietary requirements and benefits these intakes give the body. However, excess consumption of certain foods, could be far more detrimental both on the short and long run.

Sodium is majorly consumed in the form of salt (Sodium chloride). When eaten in the right quantity, it serves as an electrolyte which maintains fluid balance in the body however, when taken in excess, it pulls too much fluid, leading to high blood pressure. The recommended salt intake therefore, is less than 2,300mg daily.

Saturated fat when consumed, is a good source for the body to convert diet to energy, keeping the blood sugar levels constant. Lard, fatty meats are examples of saturated fat. However, when consumed in large amounts, saturated fats can cause cholesterol build up in blood vessels, increasing the risk of heart disease. Its intake should therefore be limited to less than 10% of daily kilocalories.

Added sugars are sources of energy to the body however, when taken in large amounts, could impede normal metabolism in the body. Added sugars should be restricted to less than 5% of daily kilocalories in diet.

Alcohol intake such as in wine, when done in moderation, has been shown to reduce the risk of heart diseases. However, when taken in excess, tolerance builds and exposes the body to the risk of conditions such as liver disease, alcohol use disorder.

4 0

4 years ago

If a patient\'s blood pressure is 145 over 65 mmHg, what is it in atmospheres (atm)

kicyunya [14]

Converting mmHg to atm is solved by division.

Example: Convert 745.0 to atm.

Solution- divide the mmHg value by the 760.0 mmHg / atm.

745 mmHg over 760.0 mmHg/atm

atm value is 0.980263

Now, I am a medical student and we have never had to convert a BP (blood pressure) to atm from mmHg, only ever kPA. SO, I am going to take a guess here and say that when you do the work to solve this, you are going to convert the Systolic (upper #) which is the 145. You should get 0.190789 and then convert the Diastolic (lower #) which is 65. You should get 0.08552632.

So your fraction so to speak should read, 0.190789/0.08552632 or 0.190789 over 0.08552632

(Just to note that is way to low of a BP, although it is irrelevant) Best wishes and good luck. "Remember, never just look for the right answer, look for why it is the right answer!"

7 0

3 years ago

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Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

2)

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

7 0

4 years ago