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3 years ago

Como se ordena la palabra ecrfinaine

Chemistry

1 answer:

labwork [276]3 years ago

5 0

Inferencia Es la palabra en orden

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A ring costs $36 more than a bracelet. the cost of the bracelet is 4/7 the cost of the ring. find the total of the two items.

avanturin [10]

So let's use some equations to represent the data [let R= cost of ring & B= cost of bracelet]

R= B + $ 36 .... (1)

B= $\frac{4}{7}$ × R ... (2)

By using simultaneous equations to solve for B and R.
Substitute eq. (1) into eq. (2)

B = $\frac{4}{7}$ × (B + $36)

B = $\frac{4}{7}$ B + $\frac{144}{7}$

$( \frac{7}{7} - \frac{4}{7} ) B = \frac{144}{7}$

$\frac{3}{7} B = \frac{144}{7}$

⇒ B = $48

By substituting value of B into ea (1)

If R = B + $36

R = ($48) + $36

= $84

∴ <span> the total of the two items = R + B
= $84 + $48
</span> = $132

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3 years ago

there are 28L of soup in a pot. Marshall serves 400mL in each bowl. If he fills 16 bowls how much soup is left in the pot? in li

Ne4ueva [31]

Total in pot=28 L
400 mL in each bowl
16 bowls filled
1000mL=1L
16 bowls(400mL/1 bowl)=6400mL
6400mL(1L/1000mL)=6.4L
28L-6.4L=21.6 L

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Which of the following is the final stage of a low mass star? *<br> 1 point

Nady [450]

You are actually giving 5 points. Ps I just got five points for telling you that

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Melting glaciers are causing _____________ to rise around the world.

Dovator [93]

Sea levels sea levels

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Find the pH. What are the pH values for the following solutions? (a) 0.1 M HCl (b) 0.1 M NaOH (c) 0.05 M HCl (d) 0.05 M NaOH

slega [8]

Answer:

(a) $pH=1$

(b) $pH=1.3$

(d) $pH=12.7$

Explanation:

Hello,

In this case, we define the pH in terms of the concentration of hydronium ions as:

$pH=-log([H^+])$

Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:

(a)

$[H^+]=[HCl]=0.1M$

$pH=-log(0.1)=1$

(b)

$[H^+]=[HCl]=0.05M$

$pH=-log(0.05)=1.3$

Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:

$[OH^-]=[NaOH]$

$pOH=-log([OH^-])$

$pH=14-pOH$

Thus, we have:

(b)

$pOH=-log(0.1)=1\\pH=14-1=13$

(d)

$pOH=-log(0.05)=1.3\\pH=14-1.3=12.7$

Best regards.

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3 years ago