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sladkih [1.3K]
3 years ago
7

234+34.1 add and round to 2 significant figures

Chemistry
1 answer:
marin [14]3 years ago
3 0
Like this? 234+34.1= 268.1 then round. If it is less than 5 then you round down if it is more then you round up. Because it is less the final number would be 268.1=268
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_2. The temperature of a 100. mL sample of gas is increased while the volume it occupies decreases. The
yaroslaw [1]

Answer

the answer is b bro

Explanation:

i did my research and got it right

5 0
2 years ago
If the rate law for a reaction A → P is rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122 M, calcula
DedPeter [7]

Rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122M.

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3 0
2 years ago
An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h
vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base  =  w

Height of the pyramid = h

Volume of the pyramid = V=\frac{1}{3}lwh

We have:

Rate at which water is filled in cube = \frac{dV}{dt}= 45 cm^3/s

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

V=\frac{1}{3}\times l^2\times h

\frac{l}{h}=\frac{6}{13}

l=\frac{6h}{13}

V=\frac{1}{3}\times (\frac{6h}{13})^2\times h

V=\frac{12}{169}h^3

Differentiating V with respect to dt:

\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}

\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}

Putting, h = 4 cm

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}

=13.20 cm/s

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0
3 years ago
What is the mass of a 0.23 L substance, if it has density of 1.90 g/mL?
jolli1 [7]

Answer:

437 g

Explanation:

as mass = density × volume

8 0
2 years ago
A patient needs to be given exactly 1000 ml of a 5.0% (w/v) intravenous glucose solution. the stock solution is 55% (w/v). how m
LiRa [457]
The patient needs 1000 ml of 5% (w/v) glucose solution
i.e  1000 ml x 5 g/ 100 ml 
where the stock solution is 55% (w/v) = 55 g / 100 ml  
So, 1000 ml x 5 g / 100 ml = V (ml) x 55 g / 100 ml 
V = 1000 x (5 / 100) / (55 / 100) = 5000 / 55 = 90.9 ml 
∴ the patient needs 90.9 ml of 55% (w/v) glucose solution

6 0
3 years ago
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