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2 years ago

Please help!!!!!!!!!!!!!!!!!!!

Chemistry

1 answer:

zimovet [89]2 years ago

3 0

Answer:

Your body cells use the oxygen you breathe to get energy from the food you eat. This process is called cellular respiration. During cellular respiration the cell uses oxygen to break down sugar. Breaking down sugar produces the energy your body needs.

Explanation:

Brainly pls

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When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200

GaryK [48]

Answer:

$2.023 m^3$ is the total displacement (volume) of the submarine.

Explanation:

Mass of water carried by submarine at 1000 ft depth = m = 2100 kg

The density of seawater at 1000 ft depth = d = $1033 kg/m^3$

Volume of the water displaced = V= ?

Total displacement of the submarine = Volume of the water displaced = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3$

$2.023 m^3$ is the total displacement (volume) of the submarine.

6 0

2 years ago

By what process do electromagnetic waves release from the materials they contact?

Rudik [331]

Answer:

The photoelectric effect

Explanation: hope it helps

5 0

3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

What mass of solute is contained in 25.4 ml of a 1.56 m potassium bromide solution?

mash [69]

Answer is: mass of <span>potassium bromide is 4.71 grams.
V(KBr) = 25.4 mL </span>÷ 1000 mL/L = 0.0254 L, volume of solution.
c(KBr) = 1.56 mol/L.
n(KBr) = c(KBr) · V(KBr).
n(KBr) = 1.56 mol/L 0.054 L.
n(KBr) = 0.0396 mol, amount of substance.
m(KBr) = n(KBr) · M(KBr).
m(KBr) = 0.0396 mol · 119 g/mol.
m(KBr) = 4.71 g.
M - molar mass.

3 0

2 years ago

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Where would sound travel the slowest

viva [34]

Answer:

Wood

Explanation:

It is because sound is a longitudinal wave

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3 years ago