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pantera1 [17]
3 years ago
5

Why might kinneys Decrease urine production in someone who is sweating a lot on a hot day?

Chemistry
1 answer:
Oduvanchick [21]3 years ago
3 0

Answer:

During exercise and heat stress, both glomerular filtration and renal blood flow are markedly reduced, resulting in decreased urine output. Therefore, when fluids are over consumed during exercise (hyperhydration), there may be a reduced ability to produce urine to excrete the excess volume

Explanation: On an average temperature day, your body will usually produce about 1 to 2 quarts of urine. Cold weather will result in your body needing to filter more blood than normal as blood is rushing to your vital organs at a higher frequency. This results in you having to pee more often.

It is concluded profuse sweating exercise results in a decrease of urinary uric acid excretion amounts and leads to increased serum uric acid after the exercise

this is some information i have to help you

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8. Ascorbic acid (Vitamin C) cannot be stored in the body, so it must be supplied in the diet. It is a C, H, O compound. If 7.75
spin [16.1K]

Answer:

Molecular formula = C6H8O6

The molar mass of vitamin C = 176.14 g/mol

Explanation:

Step 1: Data given

Mass of vitamin C = 7.75 grams

Mass of CO2 = 11.62 grams

Mass of H2O = 3.17 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 11.62 grams /44.01 g/mol

Moles CO2 = 0.264 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.264 moles CO2 we have 0.264 moles

Step 4: Calculate mass C

Mass C = 0.264 moles * 12.01 g/mol

Mass C = 3.17 grams

Step 5: Calculate moles H2O

Moles H2O = 3.17 grams / 18.02 g/mol

Moles H2O = 0.176 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.176 moles H2O we have 2*0.176 = 0.352 moles

Step 6: Calculate mass H

Mass H = 0.352 moles * 1.01 g/mol

Mass H = 0.356 grams

Step 7: Calculate mass O

Mass O = 7.75 grams - 3.17 - 0.356

Mass O = 4.224 grams

Step 8: Calculate moles O

Mass O = 4.224 grams / 16.0 g/mol

Mass O = 0.264 moles

Step 9: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.264 moles / 0.264 moles = 1

H: 0.352 moles / 0.264 moles = 1.33

O: 0.264 moles / 0.264 moles = 1

This means for each C atom we have 1.33 H atoms and 1 O atom

OR

For every 3 C atoms we have 4 H atoms and 3 O atoms

The empirical formula is C3H4O3

Step 10: Calculate the molecular formula

The molecular mass of the empirical formula is 88.07

Since the molar mass of vitamin C is between 150 and 200 g/mol

We have to multiply the empirical formula by 2

Molecular formula = 2*(C3H4O3) = C6H8O6

The molar mass of vitamin C = 176.14 g/mol

8 0
3 years ago
Which of the following is a sign that a valley glacier has moved through an area? *
Dmitry_Shevchenko [17]
B. The rocks carried by the glacier would leave the marks
7 0
4 years ago
At a festival, spherical balloons with a radius of 170.cm are to be inflated with hot air and released. The air at the festival
MissTica

Answer:

The correct answer is 1.5 balloons.

Explanation:

Based on the given question, the radius of the spherical balloons is 170 cm. Therefore, the volume of each balloon will be,  

= 4/3 × π × (170)³ cm³

= 20.5176 × 10⁶ cm³  

= 20.5176 m³

The density of air at 100 degree C s 0.946 Kg m⁻³

The mass of air in each balloon can be calculated by using the formula,  

Mass = density × volume

Mass = 0.946 Kg m⁻³ × 20.5176 m³

Mass = 19.410 Kg

The heat energy, that is, required to bring the air from 25 degree C to 100 degree C will be,  

= 19.410 × 10³ g × (100 -25) degree C × 1.009 J/g degree C

= 14.68 × 10⁵ J

The concentration of butane given is 1.00 Kg or 1000 g

The molecular weight of butane is 58.12 g per mole

The moles or n can be calculated by using the formula,  

n = mass / mol.wt

n = 1000 g/58.12 g/mol = 17.20 mol

The formation of enthalpy of butane at 25 degree C is 125.7 × 10³ J/mol. The evolution of heat energy that take place at the time of burning 17.20 moles of butane is,  

= 125.7 × 10³ J/mol × 17.20 mol

= 2.16 × 10⁶ J

The number of balloons that can be inflated with hot air is,  

= 2.16 × 10⁶ J / 14.68 × 10⁵ J per each balloon

= 1.5 balloons

Hence, maximum of 1.5 balloons can be inflated.  

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4 years ago
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A pure substance that cannot be separated into simpler substances by physical or chemical means. A pure substance composed of two or more elements that are chemically combined.
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3 years ago
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The deer carcasses lay on the side of the road on the highway.
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