Answer:
to launch aerials and also causes the explosions necessary for special effects like noise or colored light.
Explanation:
Answer: The substance is neutral
Explanation:
If you place red litmus paper into a basic substance, it turn's blue. When it comes in contact with an acidic or neutral substance, it doesn't change colour it remains red.
Blue litmus paper in acid turns red when placed in a basic or neutral solution it remains blue. From this it can be concluded that the solution is neutral.
" There will be a net movement of oxygen from outside the cell to inside the cell " Statement is True.
Explanation:
The partial pressure for oxygen in alveoli is greater under normal circumstances, and oxygen moves neatly into the blood. In addition, the partial carbon dioxide pressure throughout the blood usually is higher, such that carbon dioxide migrate clearly into the alveoli.
The few common molecules which can traverse the cell membrane by absorption (or diffusion of a sort recognized as osmosis) are water, carbon dioxide and oxygen. Metabolism is typically oxygen-needed, which is lowest in the cell within the animal and plant, so that net oxygen flows to the cell.
Answer:
Volume = 1222.5cm³
Explanation:
<em>If the question is about the volume of the rectangle:</em>
The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.
In the problem, the length of the rectangle is 0.162m = 16.2cm
The width is 7.7cm
And the height is 9.8cm
The volume is:
Volume = 16.2cm*7.7cm*9.8cm
<h3>Volume = 1222.5cm³</h3>
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
