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suter [353]
3 years ago
7

Jeepers! You scared the life out of me! change into declarative sentence​

Chemistry
1 answer:
Sedbober [7]3 years ago
7 0

You scared the life out of me!

You is the subject,

Scared is the predicate,

<u>!</u> this sign is the punctuation mark.

You might be interested in
What are the basis of making symbols????​
telo118 [61]

Answer:

1/ Using the first letter of the name of the elements.

2/  If the name of two or more elements begins with the same letter,the second letter of their name is also taken with first letter.

3/ Symbols are made by using the letters of the name of elements in other languages also.

4 0
4 years ago
The rate constant for this zero‑order reaction is 0.0240 M ⋅ s − 1 0.0240 M·s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How
maw [93]

Answer: It will take for the concentration of A to decrease from 0.900 M to 0.210 M is 28.75 s

Explanation:

Zero order reaction : A reaction is said to be of zero order if the rate is independent of the concentration of the reactants, that means the rate is directly proportional to the zeroth power of the concentration of the reactants.

Expression for the zero order kinetics:

[A]=-kt+[A]_o

where [A] = concentration left after time t = 0.210 M

[A]_o = initial concentration = 0.900 M

k= rate constant =0.0240Ms^{-1}

t = time for reaction = ?

0.210=-0.0240\times t+0.900

t=28.75s

Thus it will take for the concentration of A to decrease from 0.900 M to 0.210 M is 28.75 s

7 0
3 years ago
What is the mass of 3.5 l of co2(g)? (density = 0.0019 g/ml, answer to hundredths place)?
monitta
Density is defined as mass per unit volume.
density =  \frac{mass}{volume}
the mass of 1 mL of CO₂ is  - 0.0019 g
3.5 L = 3500 mL 
then the mass of 3500 mL of CO₂ is - 0.0019 g/mL x 3500 mL = 6.65 g
mass of 3.5 L of CO₂ is 6.65 g
4 0
4 years ago
Read 2 more answers
What are the boiling points and freezing points (in oC) of a solution of 50.3 g of I2 in 350 g of chloroform? The kb = 3.63 oC/m
patriot [66]

Answer:

Boiling point: 63.3°C

Freezing point: -66.2°C.

Explanation:

The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:

<em>Boiling point increasing:</em>

ΔT = kb*m*i

<em>Freezing point depression:</em>

ΔT = kf*m*i

ΔT are the °C that change boiling or freezing point.

m is molality of the solution (moles / kg)

And i is Van't Hoff factor (1 for I₂ in chloroform)

Molality of 50.3g of I₂ in 350g of chloroform is:

50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:

0.198 moles / 0.350kg = 0.566m

Replacing:

<em>Boiling point:</em>

ΔT = kb*m*i

ΔT = 3.63°C/m*0.566m*1

ΔT = 2.1°C

As boiling point of pure substance is 61.2°C, boiling point of the solution is:

61.2°C + 2.1°C = 63.3°C

<em>Freezing point:</em>

ΔT = kf*m*i

ΔT = 4.70°C/m*0.566m*1

ΔT = 2.7°C

As freezing point is -63.5°C, the freezing point of the solution is:

-63.5°C - 2.7°C = -66.2°C

7 0
3 years ago
How many moles of NaCl will react completely with 18.5 L F2 gas at 300.0 K and 1.00 atm?
meriva
Hello!

Data:

P (pressure) = 1 atm
V (volume) = 18.5 L
T (temperature) = 300 K
n (number of mols) = ? (in mol)
R (Gas constant) = 0.082 (atm*L/mol*K)

Apply the data to the Clapeyron equation (ideal gas equation), see:
P*V = n*R*T

1*18.5 = n*0.082*300

18.5 = 24.6n

24.6n = 18.5

n =  \dfrac{18.5}{24.6}

\boxed{\boxed{n \approx 0,752\:mol}}\end{array}}\qquad\checkmark

Note: If the feedback is to be considered, the closest response is 0.751 mol Nacl

_________________
_________________

I hope this helps. =)


3 0
3 years ago
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