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3 years ago

Describe how urine is removed from the blood and then eliminated from the body

1 answer:

7 0

Passes through the kidneys and the good stuff is kept, and the other stuff is passed through leading to the bladder. The once the bladder is full enough to send the message to the brain, urine (waste) is eliminated from the body.

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An irreversible change would be?

klemol [59]

Answer:

A. Chipping ice to flakes

Explanation:

You can't reverse the ice becoming flakes. They will stay like that until they melt

8 0

2 years ago

Calculate the mass of a substance which has a volume of 35

makkiz [27]

The answer is 17.5kg.

To get the mass of an object you do Volume × Density. The SI unit of Mass is "kg."

7 0

2 years ago

How many calories is required to change the temperature of 2.18g of water from 15.3°C to 69.5°C. The specific heat of liquid wat

lozanna [386]

The number of calories that are required to change the temperature of 2.18 g of water from 15.3 c to 69.5 c is 118.16 cal

calculation

Heat in calories = MCΔ T where,

M(mass)= 2.18 g
C(specific heat capacity)= 1.00 cal/g/c
ΔT( change in temperature)= 69.5- 15.3 =54.2 c

heat is therefore= 2.18 g x 1.00 cal/g/c x 54.2 c=118.16 cal

7 0

3 years ago

Read 2 more answers

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

3 years ago

HELP ASAP!!!!!!

Sav [38]

1, 4, 1, 2, 1

Explanation:

We have the following chemical reaction:

MnO₂ + HCl → MnCl₂ + H₂O + Cl₂

To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.

So the balanced chemical equation is:

MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂

Learn more about:

balancing chemical equations

brainly.com/question/14112113

#learnwithBrainly

7 0

3 years ago