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iris [78.8K]
3 years ago
8

A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/s2. If the cart has a beginning speed of 2.0 m/s, wh

at is its final speed?
Chemistry
1 answer:
Digiron [165]3 years ago
6 0

Answer:

22m/s

Explanation:

Given parameters:

Time  = 5s

Acceleration  = 4m/s²

Initial velocity  = 2m/s

Unknown

Final speed  = ?

Solution:

To solve this problem, we use the expression below;

         v = u + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

 So, insert the parameters and solve;

         v  = 2 + 4 x 5  = 22m/s

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The gasses in a hair spray can are at temperature 300k and a pressure of 30 atm, it
Sergeeva-Olga [200]

Answer:

900 K

Explanation:

Recall the ideal gas law:

\displaystyle PV = nRT

Because only pressure and temperature is changing, we can rearrange the equation as follows:
\displaystyle \frac{P}{T} = \frac{nR}{V}

The right-hand side stays constant. Therefore:

\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}

The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.

Substitute and solve for <em>T</em>₂:

\displaystyle \begin{aligned} \frac{(30\text{ atm})}{(300\text{ K})} & = \frac{(90\text{ atm})}{T_2} \\ \\ T_2 & = 900\text{ K}\end{aligned}

Hence, the temperature must be reach 900 K.

7 0
2 years ago
(((NEED ANSWER QUICK!!!)))<br><br> Which is the stronger conjugate base, CN- or OCN-? Explain
Shkiper50 [21]

Answer:

The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.

Explanation:

Given conjugate base CN⁻ => weak acid => HCN =>  Ka =4.9 x 10⁻¹⁰

Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴

Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻

6 0
3 years ago
Compound B contains 92.31% carbon and 7.69% hydrogen. Determine the empirical formula of B​
tester [92]

Answer:C H

Explanation:

5 0
3 years ago
Read 2 more answers
71 points Chemistry tritiation please help me
Rudiy27

Answer:

Answer in the picture

Explanation:

3 0
3 years ago
A 3.24-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio
Stolb23 [73]

Answer:

a) mass of the H2CO3 produced:

given:

Mass of sample = 3.24 g

Mass of Na2CO3 obtained after decomposition = 2.19 g

Solution :

Molar mass of NaHCO3 = 84

reaction:

2NaHCO3 → Na2CO3 + H2CO3

so it is clear that 2 mole of NaHCO3 gives 1 mole of Na2CO3 and H2CO3

Now, ICE table for the reaction is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x +x +x

E 3.24/84 -2x x x

As NaHCO3 is completely decomposed so final Concentration of NaHCO3 is zero.

=> 3.24/84 -2x = 0

=> 2x = 3.24/84

=> x = 1.62/84

The new ICE table is :

NaHCO3 Na2CO3 H2CO3

I 3.24/84 0 0

C -2x = -2(1.62/84) +x = 1.62/84 +x = 1.62/84

E 0 1.62/84 1.62/84

From the above ICE table,

it is found that (1.62/84 ) moles of H2CO3 is obtained.

Since,

The molar mass of H2CO3 is 62

=> Mass of H2CO3 obtained = moles × molar mass

=> Mass of H2CO3 obtained = (1.62 /84 ) × 62

= 1.19 grams

Mass of H2CO3 experimentally :

Mass of reactants = mass of products

=> Mass of sample = mass of Na2CO3 + mass of H2CO3

=> Mass of H2CO3 = mass of sample - mass of Na2CO3

= 3.24 - 2.19 = 1.05 g

b) Experimental mass = 1.05 g

Theoretical mass = 1.19 g

Percentage yield of H2CO3 = Experimental mass × 100 / Theoretical mass

= 1.05 × 100 /1.19

= 88.23 %

6 0
2 years ago
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