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4 years ago

How many atoms are in C4H8O

Chemistry

1 answer:

zheka24 [161]4 years ago

7 0

4 átomos de carbono.
8 átomos de hidrógeno.
1 átomo de oxígeno.

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Identify the products formed in this Brønsted-Lowry reaction. HPO2−4+NO−2↽−−⇀acid+base

klemol [59]

Answer: The products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Explanation:

According to Bronsted-Lowry, acids are the species which donate hydrogen ions to another specie in a chemical reaction.

Bases are the species which accept a hydrogen ion upon chemical reaction.

For example, $HPO^{2-}_{4} + NO^{-}_{2} \rightleftharpoons HNO_{2} + PO^{2-}_{4}$

Here, the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

Thus, we can conclude that the products formed in this Bronsted-Lowry reaction are $HNO_{2}$ and $PO^{2-}_{4}$ .

5 0

3 years ago

Which form of energy can be connected closely to climate change?

Art [367]

Answer:

COAL

Explanation:

4 0

3 years ago

How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago

How does changing the concentration of the reactants change the parts of a rate law? It changes the rate, R. It changes the rate

Sidana [21]

Answer

A. It changes the rate, R

Explanation

When we change the concentration of the reactants in a chemical reaction, it affects the rate of reaction that happens in the process. Typically, the rate of reaction will decrease with time if the concentration of the reactants decreases because the reactants will be converted to products. Similarly, the rate of reaction will increase when the concentration of reactants are increased.

4 0

3 years ago

What would be the volume in millilitres of a blood sample of 2.15 microliters (ul)?

Zanzabum

2.15 x 10⁻³mL

Explanation:

Given parameter:

Volume of blood sample in uL = 2.15uL

Conversion uL → mL

micro- and milli- are both prefixes of sub-units.

liter is a unit of volume of a substance.

micro - is 10⁻⁶

milli- is of the order 10⁻³

The problem is converting from micro to milli:

if we multiply 10⁻⁶ by 10³ we would have our milli;

1000uL = 1mL

2.15uL : 2.15uL x $\frac{1mL}{1000uL}$ = 2.15 x 10⁻³mL

learn more:

Volume brainly.com/question/5055270

#learnwithBrainly

4 0

3 years ago