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allochka39001 [22]
3 years ago
9

Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas

are formed when 48.7 grams of sodium are added to water?
Chemistry
1 answer:
zysi [14]3 years ago
5 0
Convert the 48.7 grams into moles and then use the balanced equation to convert to moles of hydrogen gas. Then divide by avogadro's number<span />
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7. Halfway through a cross country meet, a
kondor19780726 [428]

Answer:

3m/s

Explanation:

Data obtained from the question include:

Initial speed (s1) = 4 m/s

Final speed (S2) = 7m/s

Change in speed (ΔS)

ΔS = s2 — s1

ΔS = 7 — 4

ΔS = 3m/s

Therefore, the change in speed is 3m/s

7 0
3 years ago
A flask with a volume of 250.0 mL contains air with a density of 1.164 g/L. What is the mass of the air contained in the flask?
satela [25.4K]

Answer:

convert 250.0 mL in Liters :250. 0 / 1000 = 0,25 LDensity = 1.240 g/LMass

Explanation:

6 0
1 year ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
2 years ago
The Henry's law constant (kH) for O2 in water at 20°C is 1.28 × 10−3 mol/(L·atm). (a) How many grams of O2 will dissolve in 4.00
Burka [1]

Answer:

Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)

Explanation:

Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P

Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm

=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.

∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>

4 0
2 years ago
DawdDAWdADdwaadadadawddawd
Ronch [10]

Answer:

yyuuuyytt

Explanation:

ggyhggggg,,,,,,,76yyygtyyhgyyggyuhu ubbububuuh7h7h

4 0
3 years ago
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