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Montano1993 [528]
3 years ago
10

What is the valence e- (outermost e-) for the following elements:

Chemistry
1 answer:
shutvik [7]3 years ago
6 0

Answer:

A options hydrogen

Explanation:

valence outermost element is hydrogen.

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The base unit of a measured liquid called volume is?
Dafna1 [17]
The answer you need is Volume.
8 0
3 years ago
An ideal gas is contained in a cylinder with a volume of 5.0x102 mL at a temperature of 30°C and a pressure of 710. Torr. The ga
Scorpion4ik [409]

Answer:

51207 torr is the new pressure of the gas

Explanation:

We can solve this question using combined gas law that states:

P1V1T2 = P2V2T1

<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>

<em> </em>

Computing the values of the problem:

P1 = 710torr

V1 = 5.0x10²mL

T1 = 273.15 + 30°C = 303.15K

P2 = ?

V2 = 25mL

T2 = 273.15 + 820°C = 1093.15K

Replacing:

710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K

3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K

P2 = 51207 torr is the new pressure of the gas

4 0
2 years ago
Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?
muminat
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba 
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
6 0
3 years ago
Read 2 more answers
using the same balanced equation, what is the number of moles of carbon dioxide produced from 3.750 moles of propane?
kondor19780726 [428]

Answer:

11.25moles of CO2

Explanation:

First, let us generate a balanced equation for the reaction of propane to produce CO2. This reaction called Combustion. It is a reaction in which propane burns in air (O2) to produce CO2 and H20. The equation is given below:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

1mole of C3H8 produced 3moles of CO2.

Therefore, 3.750 moles of C3H8 will produce = 3.750 x 3 = 11.25moles of CO2

5 0
2 years ago
The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?
vazorg [7]
C + O2= CO2
n  =  \frac{m}{mw}
n =  \frac{5.4}{12}  \\ n  = 0.45 \: mol \: of \: carbon
n =  \frac{13.6}{12 + 16 \times 2} \\ n =  \frac{13.6}{44}  \\ n = 0.31 \: mol \: of \: carbon \: dioxide
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
4 0
3 years ago
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