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3 years ago

Which statement explains why astronauts appear to bounce when they are walking on the moon?

Chemistry

2 answers:

nata0808 [166]3 years ago

6 0

Answer:

3rd

heyy

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AnnZ [28]3 years ago

6 0

Answer:

Which statement explains why astronauts appear to bounce when they are walking on the moonExplanation:

You might be interested in

In which compound is the percent by mass of oxygen greatest?

Licemer1 [7]

Answer: BeO

Explanation:

percent by mass of oxygen $=\frac{\text {mass of oxygen}}{\text {Total mass of substance}}\times 100\%$

percent by mass of oxygen in BeO $=\frac{16g}{25.01g/mol}\times 100=63.97\%$

percent by mass of oxygen in MgO $=\frac{16g}{40.30g/mol}\times 100=39.70\%$

percent by mass of oxygen in CaO $=\frac{16g}{56.08g/mol}\times 100=28.53\%$

percent by mass of oxygen in SrO $=\frac{16g}{103.62g/mol}\times 100=15.44\%$

4 0

3 years ago

Read 2 more answers

If 50 g of lead (of specific heat 0.11 kcal/kg ∙ C°) at 100°C is put into 75 g of water (of specific heat 1.0 kcal/kg ∙ C°) at 0

Contact [7]

Answer: The final temperature is $279.8K$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of lead = 50 g

$m_2$ = mass of water = 75 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of lead = $100^oC=373K$

$T_2$ = temperature of water = $0^oC=273K$

$c_1$ = specific heat of lead = $0.11kcal/kg^0C$

$c_2$ = specific heat of water= $1.0kcal/kg^0C$

Now put all the given values in equation (1), we get

$50\times 0.11\times (T_{final}-373)=-[75\times 1.0\times (T_{final}-273)]$

$T_{final}=279.8K$

Therefore, the final temperature of the mixture will be 279.8 K.

3 0

4 years ago

How many grams are there in a 14.5 moles lithium permanganate

frutty [35]

There are 1825.6 g in a 14.5 moles Lithium permanganate

<h3> Further explanation</h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

Moles can also be determined from the amount of substance mass and its molar mass :

$\tt mol=\dfrac{mass}{MW}$

moles of Lithium permanganate = 14.5

Lithium permanganate (LiMnO4) MW=125.9 g/mol, so mass :

$\tt mass=mol\times MW\\\\mass=14.5\times 125.9~g/mol\\\\mass=1825.6~g$

5 0

3 years ago

2K + 2HBr → 2 KBr + H2

Inessa [10]

Answer:

$\large \boxed{\text{0.0503 g}}$

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 39.10 80.41 2.016

2K + 2HBr ⟶ 2KBr + H₂

m/g: 5.5 4.04

a) Limiting reactant

(i) Calculate the moles of each reactant

$\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}$

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:

The molar ratio of H₂:K is 1:2.

$\text{Moles of H}_{2} = \text{0.141 mol K} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol K}} = \text{0.0703 mol H}_{2}$

From HBr:

The molar ratio of H₂:HBr is 3:2.

$\text{Moles of H}_{2} = \text{0.049.93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol HBr}} = \text{0.024 97 mol H}_{2}$

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

$\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\ \text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$ }$

3 0

3 years ago

C8H18+ O2> CO2+ H2O how to make this balanced?

Sonbull [250]

Product:
C: 1
O:3
H: 2

reactant:
C: 8
H: 18
O: 2

first you need to put probably an eight in front of the Carbon.
Next, add a coefficient of 9 to Hydrogen
finally, add a 3 INSTEAD of a 2 to Oxygen

should look like:
C8H18+O3>C8O2+H9O

7 0

3 years ago