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3 years ago

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The FIRST space station placed into orbit around the earth followed by a team sent from earth to link up with the station was de

veloped by __________. A. Japan B. The Soviet Union C. The United States D. The European Space Agency

2 answers:

Radda [10]3 years ago

8 0

B. the soviet union .........

Marianna [84]3 years ago

4 0

The FIRST space station placed into orbit around the earth followed by a team sent from earth to link up with the station was developed by <span>The Soviet Union. The correct option among all the options that are given in the question is the second option or option "B". I hope the answer has come to your help.</span>

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Two samples of aluminum foil, 27 grams and 54 grams, are cut out from a roll. Which of the following properties is different for

Delicious77 [7]

Before answering this question, you must know the concept between extensive and intensive property. The extensive property does not depend on the amount of substance (like mass), which is the opposite of intensive properties. From the given choices, the rest are extensive properties except for <em>amount of matter</em>. Hence, that is the answer.

4 0

4 years ago

Read 2 more answers

The percentage of fe in an iron ii sample was determined by redox titration with cr2o7 calculate the molarity of the standard k2

Archy [21]

The molarity is 0.018M and the percentage is 8.46%.

Net ionic reaction is

$\mathrm{Cr} 2 \mathrm{O}^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \rightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H} 2 \mathrm{O}$

$1 \mathrm{~mol}$ of $\mathrm{Cr} 2 \mathrm{O} 7(-2)$ reacts with $6 \mathrm{~mol}$ of $\mathrm{Fe}(2+)$ to form $6 \mathrm{~mol}$ of $\mathrm{Fe}(3+)$ and $2 \mathrm{~mol}$ of $\mathrm{Cr}(3+)$

Find molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

Molarity = moles of $\mathrm{K} 2 \mathrm{Cr} 207 /$ Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7(\mathrm{~L}$ )

Moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ = grams of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ molar mass of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

$=1.2275 \mathrm{~g} / 294.19 \mathrm{~mol}=0.0042 \mathrm{~g} / \mathrm{mol}$

Molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7=$ moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ / Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ (L)

$=0.0042 \mathrm{~g} / \mathrm{mol} / 0.250 \mathrm{~L}=0.017 \mathrm{M}$

Find molarity of $\mathrm{Fe}$ (II)

Molarity of $\mathrm{Fe}(\mathrm{II})=6 \times$ molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 \times$ volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ volume of $\mathrm{Fe}$ (II)

$=6 \times 0.017 \mathrm{M} \times 0.03598 \mathrm{~L} / 0.200 \mathrm{~L}$

$=0.018 \mathrm{M}$

Find moles of $\mathrm{Fe}(\mathrm{II})$

Moles of $\mathrm{Fe}($ II) =$ molarity of $\mathrm{Fe}$ (II) $\times$ volume of $\mathrm{Fe}$ (II)

$=0.018 \mathrm{M} \times 0.200 \mathrm{~L}=0.004 \mathrm{~mol}$

Find mass of Fe(II)

Mass of $\mathrm{Fe}( II )=$ moles of $\mathrm{Fe}( II$ ) $\times$ molar mass of $\mathrm{Fe}(\mathrm{II})$

$=0.004 \mathrm{~mol} \times 55.85 \mathrm{~g} / \mathrm{mol}$

$=0.205 \mathrm{~g}$

Find $\% \mathrm{Fe}(\mathrm{II})$ in unknown sample

$\begin{aligned}\% \mathrm{Fe} &=(\text { mass of } \mathrm{Fe} / \text { weight of unknown sample }) \times 100 \\&=(0.205 \mathrm{~g} / 2.4234 \mathrm{~g}) \times 100 \\&=8.46 \%\end{aligned}$

Molarity is the concentration of a solution expressed as the number of moles of solute dissolved in each liter of solution.

concentration is the amount of a substance per defined space. Concentration usually is expressed in terms of mass per unit volume.

To know more about MOLARITY visit : brainly.com/question/8732513

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7 0

2 years ago

Will give brainliest

Nostrana [21]

Answer:

I think 18 not entirely sure though

Explanation:

Group 18 are the noble group so 16 + 2 = 18

5 0

3 years ago

Read 2 more answers

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such li

lozanna [386]

Answer: 4.50*10^-6T (0.00000450071T)

Explanation: A current carrying conductor has been knowing to generate a specific amount of magnetic field.

This is given by the Bio-savart law (mathematical).

The Bio-savart law is a mathematical equation that gives the value of strength of the magnetic field created by a current carrying conductor.

B=(Uo* I) /2πr

Where

B= strength of magnetic field

Uo = magnetic permeability in free space = 1.257 *10^-6

r = distance between current carrying conductor and any reference point.

By doing the neccesary algebra, we have

B=(1.257 *10^-6 * 180)/ (2 * 3.142 * 8)

B= 2.2626 *10^-4 / 50.2857

B=4.5 * 10^-6T (0.00000450071T)

6 0

3 years ago

A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon

Anettt [7]

First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed $v_x=30 m/s$ , and an accelerated motion on the y-axis, with initial speed $v_y=20 m/s$ and acceleration $g=9.81 m/s^2$ :
$S_x(t)=v_xt$
$S_y(t)=v_y t- \frac{1}{2} gt^2$
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
$S_y(t)=0$
Therefore:
$v_y t - \frac{1}{2}gt^2=0$
which has two solutions:
$t=0$ is the time of the beginning of the motion,
$t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s$ is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
$S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m$

4 0

3 years ago