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3 years ago

9

When acid reacts with a base, what compounds are formed? (Straight answer plz)

1 answer:

True [87]3 years ago

5 0

Salt and Water is the correct answer.

hope this helps!

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What mass of aluminum (in g) would be

antiseptic1488 [7]

Mass of Aluminum (in g) : 4.104

<h3>Further explanation</h3>

Reaction

2 Al(s) + 6 HCl(aq) → 2 AICI₃ (aq) + 3 H₂(g)

mol HCl :

$\tt 1.3\times 0.35=0.455$

mol Al :

$\tt \dfrac{2}{6}\times 0.455=0.152$

mass of Aluminum :

$\tt 0.152\times 27~g/mol=4.104~g$

5 0

3 years ago

A small sack of sand has a density of 1.5 g/cm3 and a mass of 1500 g. How much space (volume) does the sand occupy?

d1i1m1o1n [39]

The answer would be 1000 cm3

8 0

3 years ago

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In the Cannizzaro reaction of benzaldehyde ( aka solventless disproportionation of benzaldehyde in the presences of a strong bas

choli [55]

Answer:

Reaction products are shown below

Explanation:

In cannizaro reaction, benzaldehyde produces benzoic acid and benzyl alcohol through disproportion.
Here, a hydroxyl group first gives a nucleophilc addition at carbonyl center of benzaldehyde.
Then a hydride transfer takes place to another molecule of benzaldehyde
Thus, benzoic acid and benzyl alcohol are produced.
Full reaction mechanism has been shown below

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4 years ago

HELPPPPP WILL MARK BRAINLIEST

ivolga24 [154]

Answer: a.b.d

Explanation:

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3 years ago

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During a routine check of the fluoride content of Gotham City\'s water supply, the following results were obtained from replicat

Aleksandr-060686 [28]

Answer:

The mean is $x=0.503\frac{mg}{L}$

The 90% confidence interval is:

$i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}]$

Explanation:

1. First organize the data:

$x_{1}=0.487$

$x_{2}=0.487$

$x_{3}=0.511$

$x_{4}=0.511$

$x_{5}=0.519$

As there are 5 data, the sample size (n) is n=5

2. Calculate the mean x:

The mean is calculated adding up all the data and divide them between the sample size.

$x=\frac{0.511+0.487+0.511+0.487+0.519}{5}$

$x=0.503\frac{mg}{L}$

3. Find 90% confidence interval.

The formula to find the confidence interval is:

$i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})]$ (Eq.1)

where x is the mean, d is the standard deviation and n is the sample size.

And

$1-\alpha=0.90$

$\alpha=0.10$

$\frac{\alpha}{2}=0.05$

$z_{0.05}=1.645$

4. Find the standard deviation

$d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}$

$d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}$

$d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}$

$d=\sqrt{2.24*10^{-4}}$

$d=0.015$

5. Replace values in (Eq.1):

$i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})]$

For the addition:

$i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.514$

For the subtraction:

$i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.492$

The 90% confidence interval is:

$i_{0.90}=[0.492,0.514]$

4 0

3 years ago