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2 years ago

A sky rocket is launched with a force of 10 N and accelerates at 20 m/s squared what is the mass of this skyrocket

Chemistry

1 answer:

tia_tia [17]2 years ago

4 0

Answer: 0.5 kg

Explanation:

Force of launched skyrocket = 10 N

Acceleration of skyrocket = 20 m/s squared (i.e 20m/s^2)

Mass of the skyrocket = ?

Recall that Force is the product of the mass of an object by the acceleration by which it moves.

i.e Force = Mass x Acceleration

10N = Mass x 20m/s^2

Mass = (10N/20m/s^2)

Mass = 0.5 kg

Thus, the mass of this skyrocket is 0.5 kilograms

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If the density of nitrogen gas at a certain pressure and temperature is 1.20 g/l, how many moles of nitrogen gas are in 15.0 l

Daniel [21]

Density of nitrogen= 1.20 g/l
no. of grams of nitrogen=m= (1.20g/l)(15.0l)= 18 g
molar mass of nitrogen=M= (14*2)= 28 g/mol

no. of moles of nitrogen=m/M=18/28= 0.643 moles

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2 years ago

How many particles are in 4g of NaCl? (1mol of NaCl = 58g) – 2 step conversion

oksian1 [2.3K]

Answer:

4.16 * 10²⁴ particles

Explanation:

1. We convert from grams to moles

4 g NaCl / 58 g of NaCl = 0.0689655172 moles

-> 0.0690 moles

2. Now we will convert from moles to particles by multiplying by avogadro's number..

0.0690 * 6.022 *10²³ = 4.15518 * 10²⁴ particles or 4.16 * 10²⁴ particles

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2 years ago

ILL GIVE U 100 POINTS!!!!

Alexxx [7]

✿I'll take that 50 pts ✿

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hope it helps ❤~

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2 years ago

Read 2 more answers

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

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2 years ago

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In the engineering design process what do engineers do immediately after testing a prototype

mote1985 [20]

Answer:

Iterate to improve the solution

Explanation:

apex

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2 years ago