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gladu [14]
3 years ago
7

Please help. Im not a very smart person

Chemistry
1 answer:
GenaCL600 [577]3 years ago
7 0
Either it’s, it is released when the reaction is complete or it is changed into atoms of carbon and oxygen during the reaction
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Consider 2H2 + O2 → 2H2O. To produce 1.2 g water, how many grams of H2 are required? Report to the correct number of significant
Elden [556K]

Answer:

0.133 mol (corrected to 3 sig.fig)

Explanation:

Take the atomic mass of H=1.0, and O=16.0,

no. of moles = mass / molar mass

so no. of moles of H2O produced = 1.2 / (1.0x2+16.0)

= 0.0666666 mol

From the equation, the mole ratio of H2:H2O = 2:2 = 1:1,

meaning every 1 mole of H2 reacted gives out 1 mole of water.

So, the no, of moles of H2 required should equal to the no, of moles of H2O produced, which is also  0.0666666 moles.

mass = no. of moles x molar mass

hence,

mass of H2 required = 0.066666666 x (1.0x2)

= 0.133 mol (corrected to 3 sig.fig)

3 0
3 years ago
Convert the pressure 525.4 torr to kPa.
Virty [35]

Answer:

A. 70.0 kpa

Explanation:

Given

Pressure = 525.4 torr

Required

Convert to kPa

From standard unit of conversion;

1\ torr = 0.133\ kPa

Multiply both sides by 525.4

525.4 * 1\ torr = 525.4 * 0.133\ kpa

525.4\ torr = 70.047572\ kpa

525.4\ torr = 70.0\ kpa <em>(Approximated)</em>

<em>From the list of given options, option A is the closest answer after 525.4 torr is converted to its kpa equivalent</em>

8 0
3 years ago
Read 2 more answers
8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c
dangina [55]

Answer:

Partial pressure of O_{2} in the gas was 733 torr and mass of KClO_{3} in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of O_{2})

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of O_{2}= (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that O_{2} behaves ideally. Hence-

                                            PV=nRT

where P is pressure of O_{2}, V is volume of O_{2} , n is number of moles of O_{2} , R is gas constant and T is temperature in kelvin

here P = 733 torr = (733\times 0.001316)atm = 0.9646 atm

        V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

   So, n=\frac{PV}{RT}

                   = \frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}

                   = 0.0259 moles

As 3 moles of O_{2} are produced from 2 moles of KClO_{3} therefore 0.0259 moles of O_{2} are produced from (\frac{2\times 0.0259}{3}) moles or 0.0173 moles of KClO_{3}.

Molar mass of KClO_{3}= 122.55 g

So mass of KClO_{3} in sample = (0.0173\times 122.55)g

                                                                    = 2.12 g

7 0
3 years ago
A scientist took notes about the fossil in the picture.
Dimas [21]

Answer:

b. forest ....

I think ansesters also lived in forest by wearing leaves like their clothes .... etc

6 0
3 years ago
How many atoms are there in 90.2 g of selenium?
tia_tia [17]

Answer:First calculate the moles of selenium by multiplying the mass x molar mass 90.2 × 78.96 = 7122.2 moles. Number of moles multiplied by ...

Explanation:

3 0
3 years ago
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