All categories

3 years ago

The muscles in the stomach aid in digestion by __________.

Chemistry

2 answers:

dsp733 years ago

4 0

Churning or mixing food.

The stomach is a muscular sac that lies between theesophagus and thesmall intestine in the upper abdomen. The stomach is not the only part of your digestive system that absorbs food but rather is a part of the digestive system and important for churning food into a consistency that is easier to digest for the rest of your intestines.

Yanka [14]3 years ago

4 0

Answer:

B.churning, or mixing, food

Explanation:

got it right on the quiz . :)

You might be interested in

A thin sheet of aluminum can block _____.

BartSMP [9]

It can block light. Please mark Brainliest!!!

6 0

3 years ago

In one scene in the movie The Godfather II,

Ad libitum [116K]

10 cm in a cube to find volume is 10 by 10 by 10
1000 cm cubed

10 cm is 0.1 m
0.1 times 0.1 times 0.1 is 0.001 metres cubed
19,300 times 0.001 is
19.3 kg
approximately 193newtons
yes it could be passed around casually

5 0

3 years ago

What is the most metallic item on the pterotic table

statuscvo [17]

Answer:

francium or cesium

7 0

3 years ago

Read 2 more answers

Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen

OLga [1]

Answer:

5.9 kg

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ: 55.84

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

m/kg: 7.0

(a) Moles of Fe

$\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}$

(b) Moles of CO

$\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}$

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

$\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}$

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ: 32.00

2C + O₂ ⟶ 2CO

n/mol: 261

(a) Moles of O₂

$\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}$

(b) Mass of O₂

$\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \text{4180 g O}_{2}$

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

$\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}$

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.

6 0

4 years ago

What the reaction rates chart means

Zina [86]

Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. ... The graph shows the change in the number of A and B molecules in the reaction as a function of time over a 1 min period (bottom)

7 0

4 years ago