All categories

3 years ago

A feeling of weightlessness in orbit is called

Chemistry

2 answers:

Serggg [28]3 years ago

8 0

Microgravity is your answer

Katen [24]3 years ago

6 0

A feeling of weightlessness in orbit is called microgravity.

You might be interested in

The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .

3 0

3 years ago

Type the correct answer in the box. Express your answer to three significant figures. Iron(II) chloride and sodium carbonate rea

Ad libitum [116K]

Answer:

The reaction can produce 287 grams of iron(II) carbonate

Explanation:

To solve this question we must find the moles of iron(II) chloride that react. Using the chemical equation we can find the moles of iron(II) carbonate and its mass -Molar mass FeCO3: 115.854g/mol-

<em>Moles FeCl2:</em>

1.24L * (2.00mol / L) = 2.48 moles FeCl2

As 1 mol FeCl2 produce 1 mol FeCO3, the moles of FeCO3 = 2.48 moles

2.48mol * (115.854g / mol) =

<h3>The reaction can produce 287 grams of iron(II) carbonate</h3>

5 0

2 years ago

Which type of heat island is

icang [17]

Answer:

a fast moving heat island

Explanation:

cause there is a high amount of heating molecule produced.

3 0

2 years ago

For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e

sergiy2304 [10]

Answer:a) 11.34 g of ethane $(C_2H_6)$ can be formed

b) $C_2H_4$ is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$

1. $\text{Moles of} H_2=\frac{4.21}{2}=2.10moles$

2. $\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ require 1 mole of $H_2$

Thus 0.378 moles of $C_2H_4$ will require= $\frac{1}{1}\times 0.378=0.378moles$ of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

moles of $H_2$ left = (2.10-0.378) = 1.72 moles

mass of $H_2$ left= $moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g$

According to stoichiometry :

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.378 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.378=0.378moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g$

Thus 11.34 g of ethane is formed.

4 0

3 years ago

The CO2 that builds up in the air of a submerged submarine can be removed by reacting it with sodium peroxide. 2 Na2O2 (s) + 2 C

grandymaker [24]

Answer: The sodium peroxide needed per sailor in a 24 hr period is 674 grams.

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1.02 atm

V = Volume of gas = 150.0 ml = 0.15 L

n = number of moles of carbon dioxide = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $20^0C=(20+273)K=293K$

$n=\frac{PV}{RT}$

$n=\frac{1.02atm\times 0.15L}{0.0821 L atm/K mol\times 293K}=0.006moles$

$2Na_2O_2(s)+2CO_2(g)\rightarrow 2Na_2CO_3(s)+O_2(g)$

According to stoichiometry:

2 moles of carbon dioxide require = 2 moles of sodium peroxide

Thus 0.006 moles carbon dioxide require= $\frac{2}{2}\times 0.006=0.006$ moles of sodium peroxide

Mass of sodium peroxide needed for 24 hour= $moles\times {\text {Molar mass}}=0.006mol\times 78g/mol\times 24\times 60=674g$

Thus sodium peroxide needed per sailor in a 24 hr period is 674 grams.

4 0

2 years ago