All categories

2 years ago

A pogo stick bounces 30 M in 10 secs what is the bouncers average speed?

Chemistry

2 answers:

DiKsa [7]2 years ago

7 0

I would love to help but it is 3 or is it 3 million?

AleksAgata [21]2 years ago

4 0

three i hope this helps

You might be interested in

BRAINLIEST FOR RIGHT ANSWERS PLS HELP

AVprozaik [17]

Answer:

Lower

Option B

Pls mark it as brainlist.

5 0

1 year ago

What do scientists use scientific notation for?

tamaranim1 [39]

Scientists use scientific notation to simplify numbers, basically. When dealing with really big numbers or really small numbers, the usage of scientific notation prevents them from having to write a bunch of zeroes.
Hope that helped you!

8 0

2 years ago

Read 2 more answers

29.3g CO2 and 20.8g H2O

docker41 [41]

Answer:

the most best answer is 50.1 you never find such type of answer

6 0

2 years ago

Calcium carbonate is a compound that is used to make chalk for drawing on rocks or pavement. One molecule has one atom of calciu

Black_prince [1.1K]

Answer:

the chemical formula is caca3

6 0

2 years ago

Read 2 more answers

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

7 0

2 years ago