All categories

3 years ago

The respiratory system is what brings in food and breaks it down True or false

Chemistry

1 answer:

Gala2k [10]3 years ago

5 0

Answer:

false

Explanation:

the respiratory system includes the lungs and heart

not food

You might be interested in

A neutral solution has a(n) _______ amount of hydroxide ions compared to hydrogen ions.

erastovalidia [21]

The answer is 'equal'. Hydroxide ions are OH- and Hydrogen ions are H+. Have you noticed they're opposite charges? Positive + negative = neutral. That's all there is to it :)

4 0

3 years ago

Activity In this activity, you’ll draw and count atoms to understand how nitrogen and hydrogen react to make ammonia. Begin by s

vredina [299]

Answer:

Explanation:

3 0

3 years ago

A Snickers bar contains 27.0 g of sugars (carbohydrates), 4.0 g of protein and 12.0 g of fat. How many food calories (Cal) are c

amm1812

Answer:

There are 232 calories in the bar

Explanation:

Carbohydrates and protein are both 4 calories per gram, while fat is 9 calories per gram.

$27*4=108\\4*4=16\\12*9=108\\108+108+16=232$

4 0

3 years ago

Consider the genetic cross for absent-mindedness, which is a dominant trait. What is the probability that the offspring of this

4vir4ik [10]

D)100%

Every single one has the dominate gene so all the offspring will get it

I hope that helps!

8 0

3 years ago

Read 2 more answers

Will mark the brainiest later for correct answers! Please show work.

Elodia [21]

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}$

a. moles in 14.08 g of $C_{12}H_{22}O_{11}$ = $\frac{14.08g}{342.3g/mol}=0.04113moles$

molecules in 14.08 g of $C_{12}H_{22}O_{11}$ = $0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}$

b. moles in 17.75 g of NaCl = $\frac{17.75g}{58.5g/mol}=0.3034moles$

molecules in 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

formula units 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

c. moles in 20.06 g of $CuSO_4.5H_2O$ = $\frac{20.06g}{249.68g/mol}=0.08034moles$

formula units in 20.06 g of $CuSO_4.5H_2O$ = $0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}$

7 0

3 years ago