All categories

3 years ago

When blood moves from an atrium to a ventricle, it must pass through a __________

Chemistry

2 answers:

tangare [24]3 years ago

4 0

When blood moves from an atrium to a ventricle, it must pass through a one-way valve.

coldgirl [10]3 years ago

3 0

Answer:

Right side of the heart

Explanation:

Blood enters the heart through two large veins, the inferior and superior vena cava, emptying oxygen-poor blood from the body into the right atrium of the heart. As the atrium contracts, blood flows from your right atrium into your right ventricle through the open tricuspid valve.

You might be interested in

What is 63g+9cm density

Tatiana [17]

Mass = 63 g

volume = 9 cm³

D = m / V

D = 63 / 9

D = 7.0 g/cm³

3 0

4 years ago

25. To separate benzene from water, a 850 g aqueous solution of benzene is cooled from

Sonja [21]

Answer:

132.94 kilo joules

Explanation:

We are given;

Mass of benzene solution is 850 g
Temperature change from 95°C to 3° C = 92° C or 92 K

We are required to determine the energy released by the cooling of benzene.

Taking the specific heat capacity of liquid benzene as 1.7 kJ/kg/K

Then we can determine the amount of heat energy released by using the formula;

Heat = Mass × specific heat × change in temperature

= 0.85 kg × 1.7 kJ/kg/K × 92 K

= 132.94 kilo joules

Therefore, the heat released by cooling benzene is 132.94 kilo joules

6 0

3 years ago

What is the zonecreated if force of separation occurs?

Free_Kalibri [48]

I can’t answer you but here have a meme

4 0

3 years ago

Calculate the number of atoms in 1.0 mole of O2

natali 33 [55]

Number of molecules = moles x avogadros constant so 1 x 6.02x10^23 = 6.02x10^23

O2 has two atoms therefore you times your answer by 2 making it

= 1.204x10^24

7 0

3 years ago

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the

Fofino [41]

Answer:

(a) $102.6g/mol$

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

$V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}$

Solving for the molarity of base we've got:

$M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L$

Now, we can compute the moles of the base as:

$n_{base}=0.425mol/L*0.1L=0.0425mol$

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

$\frac{4.36g}{0.0425mol} =102.6g/mol$

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

$102.6g/mol-16g/mol-1g/mol=85.6g/mol$

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.

8 0

3 years ago