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3 years ago

9

What could using cobalt glass be helpful to identify?

1 answer:

satela [25.4K]3 years ago

5 0

Answer:

Using cobalt glass could be helpful to identify elements that weakly emit blue and/or violet.

Explanation:

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How many grams are in 9.97 moles of Be(NO3)2?<br> Use two digits past the decimal for all values.

Simora [160]

Answer:

1,869.97 grams of Be(NO3)2

Explanation:

Be(NO3)2 = Be N2 O6

Be=9.012182g/mole

N2=28.0134g/mole

O6=96g/mole

therefore Be(NO3)2 gives you 187.56g in one mole

so 9.97 moles means there is 9.97 times more

9.97mole Be(NO3)2 * 187.56g Be(NO3)2/1mole Be(NO3)2 = 1,869.97g of Be(NO3)2

4 0

3 years ago

In the target above, how would you classify the "hits"? (note: an accurate "hit" is a bull's-eye.)

Dominik [7]

B) Accurate is the right answer

5 0

4 years ago

A recipe for candy requires that the liquid ingredients be stirred constantly until the liquid reaches a temperature of 140°C. W

inn [45]

Answer:

Wood

Explanation:

Wood because the heat would not be able to travel to the hand to burn it unlike the metal

8 0

2 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

3 years ago

what is the proper scientific notation for 0.000476? a.0.476 × 10-3 b.4.76 × 10-4 c.47.6 × 10-5 d.476 × 10-6

-BARSIC- [3]

B. 4.75 × 10-4 is the answer

6 0

3 years ago