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ICE Princess25 [194]
2 years ago
5

Iron + Hydrochloric acid --> ?

Physics
2 answers:
marusya05 [52]2 years ago
8 0
<h3>Iron - Fe</h3>

<h3>Hydrochloric Acid- HCl</h3>

<h2><u>Solution</u></h2>

\bold{Fe  +2HCl \rightarrow FeCl _{}{ \tiny2}  + H{ \tiny{2}}}

\therefore \bold{\fbox{{Balanced}}}

Iron + Hydrochloric Acid \rightarrow Ferrous Chloride + Hydrogen

<h2>Hope This Helps You ❤️</h2>
Paladinen [302]2 years ago
6 0

Ferrous Chloride + Hydrogen

Chemical equation: Fe + 2HCl ->FeCl2 + H2

<em>Feel free to mark as brainliest :D</em>

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Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How fa
zloy xaker [14]

Answer:

150000000000 m

0.0000005 seconds

33.33 ns

Explanation:

Speed of electromagnetic waves through vacuum = 3\times 10^8\ m/s

Echo time = 1000 seconds

Echo time is the time taken to reach the object and come back to the observer

Distance = Speed×Time

Distance=3\times 10^8\times \frac{1000}{2}=150000000000\ m

Venus is 150000000000 m away from Earth

Time = Distance / Speed

Time=\frac{75}{3\times 10^8}=0.00000025\ seconds

Echo time will be twice the time

Echo\ time=2\times 0.00000025=0.0000005\ seconds

The echo time will be 0.0000005 seconds

Difference in time = Difference in distance / Speed

\Delta t=\frac{\Delta d}{v}\\\Rightarrow \Delta t=\frac{10}{3\times 10^8}\\\Rightarrow \Delta=33.33\ ns

The accuracy by which I will be able to measure the echo time is 33.33 ns

5 0
3 years ago
HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0
3 years ago
Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210
kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light (\lambda )=565nm \approx 565\times 10^{-9}m

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

=\frac{2\lambda L}{b}

=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

=9.685 mm

(b)Width of the first order bright fringe

Y_1=\frac{\lambda \times L}{b}

Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

Y_1=4.84mm

5 0
3 years ago
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agasfer [191]

v = 2.45×10^3\:\text{m/s}

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F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

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F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

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v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

6 0
2 years ago
Displacement vectors of 4 km north, 2km south, 5 km north, and 5km south combine to a total displacement of A 2 km south B 16 km
Phantasy [73]
The north vectors add up as so the south vectors.  Then subtract the two.  For north its 4 + 5 = 9.  South is 2 + 5 = 7.   Then 9-7 = 2km North (D)
7 0
3 years ago
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