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Anestetic [448]

3 years ago

6

Hunter wants to measure the weight of a dumbbell.he should

2 answers:

kirza4 [7]3 years ago

6 0

Answer:

D. Use a scale and record the weight in N

Explanation:

Pepsi [2]3 years ago

5 0

<span>He should use a scale and record the weight in N. The beaker is used for liquid measurement and the scale is used for measuring sold things. A dumbbell is not a liquid material that is why the need of the scale will give it the accurate measurement that Hunter wants to know.</span>

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What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?

White raven [17]

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

$Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+$

Initial conc. 0.450 0 0

At eqm. (0.450-x) x x

As we are given:

$K_a=1.00\times 10^{-5}$

The expression for equilibrium constant is:

$K_a=\frac{(x)\times (x)}{(0.450-x)}$

Now put all the given values in this expression, we get:

$1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}$

$x=0.00212M$

The concentration of $H^+$ = x = 0.00212 M

Now we have to calculate the pH of solution.

$pH=-\log [H^+]$

$pH=-\log (0.00212)$

$pH=2.67$

Therefore, the pH of the solution is, 2.67

8 0

3 years ago

When one form of energy is converted to another form in any physical or chemical change, energy input alwasy equals energy outpu

aliya0001 [1]

In the context of chemistry, yes. Energy input is always equal to the energy output.

5 0

3 years ago

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

AlexFokin [52]

Explanation:

The given cell reaction is as follows.

$In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)$

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : $In(s) \rightarrow In^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode; Reduction-half reaction : $Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)$ ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

$2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}$

$3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)$

Net reaction: $2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

Thus, we can conclude that the overall cell reaction is as follows.

$2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

4 0

3 years ago

Please fill this formula

Maurinko [17]

1.) Na
2.) Cl ( at the second blank)
sodium metal+hydrochloric acid

3 0

3 years ago

HNO3 + KOH + H2O + KNO3<br><br> ACID:<br> BASE:<br> SALT:

SSSSS [86.1K]

Explanation:

acid, HNO3

base, KOH

salt ,KNO3

4 0

3 years ago