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3 years ago

If the concentration of brown sand is 2.0 ppm, how many grains of brown sand are in the bucket?

Chemistry

1 answer:

AysviL [449]3 years ago

8 0

I believe the complete given is that the bucket of sand contains 9.5 billion grains of sand. That is:

bucket = 9,500,000,000 grains

and the concentration of brown sand is 2 ppm:

concentration brown sand = 2 / 1,000,000

Therefore:

brown sand = 9,500,000,000 * (2 / 1,000,000)

<span>brown sand = 19,000 grains</span>

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Which of the following elements has the largest atomic radius? (2 points)

UNO [17]

Cobalt
hope this helps :)

5 0

3 years ago

Read 2 more answers

A 1000g Ni rod, heated to 150. °C was placed in 1.00 kg of water at 25.0 °C. The final temperature of the water was 26.3 °C. Wha

Shtirlitz [24]

Answer : The molar specific heat of Ni is, $2.576J/mole^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of Ni = ?

$c_1$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of Ni = 1000 g

$m_2$ = mass of water = 1 kg = 1000 g

$T_f$ = final temperature of water = $26.3^oC$

$T_1$ = initial temperature of Ni = $150^oC$

$T_2$ = initial temperature of water = $25^oC$

Now put all the given values in the above formula, we get

$1000g\times c_1\times (26.3-150)^oC=-1000g\times 4.18J/g^oC\times (26.3-25)^oC$

$c_1=0.0439J/g^oC$

Now we have to calculate the molar specific heat of Ni.

$\text{Molar specific heat of Ni}=\text{Specific heat of Ni}\times \text{Molar mass of Ni}=(0.0439J/g^oC)\times (58.69g/mole)=2.576J/mole^oC$

Therefore, the molar specific heat of Ni is, $2.576J/mole^oC$

4 0

3 years ago

Will mark as Brainliest.

rodikova [14]

The answer would be c

6 0

4 years ago

What volume, in mL, of 4.50 M NaOH is needed to prepare 250. mL of 0.300 M NaOH?

lisabon 2012 [21]

Answer:

16.7 mL

Explanation:

Convert 250 mL to L.

250 mL = 0.250 L

Calculate the amount of moles of NaOH in 250 mL of 0.300 M NaOH.

0.250 L × 0.300 M = 0.075 mol

Using this amount of moles, you need to find out what volume of 4.50 M will give you that many moles. You can do this by dividing the amount of moles by the molarity.

(0.075 mol)/(4.50 M) = 0.0167 L

Convert from L to mL.

0.0167 L = 16.7 mL

7 0

3 years ago

You are given an unknown mixture containing NaCl and NaHCO3. When you carry out the heating exactly as described in part A, only

SOVA2 [1]

Answer:

1.52g NaHCO3 were in the original mixture.

Mass percent: 64.1%

Explanation:

When NaHCO3 heats it descomposition occurs as follows:

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g).

The loss in mass is because of the evaporization of CO2 and H2O. As both are in the same porportion, its molar mass is the sum of both compounds (44g/mol + 18g/mol = 62g/mol)

Loss in mass: 74.80g - 74.24g = 0.56g.

In moles:

0.56g * (1mol / 62g) = 0.00903 moles of gas.

As 1 mole of the gases comes from 2 moles of NaHCO3:

<em>Moles NaHCO3:</em>

0.00903 moles of gas * (2 moles NaHCO3 / 1 mole gas) = 0.018 moles NaHCO3.

In grams (Molar mass NaHCO3: 84g/mol):

0.018 moles NaHCO3 * (84g / mol) = 1.52g NaHCO3 were in the original mixture.

The mass of the mixture was:

74.80g - 72.428g = 2.372g

That means mass percent of NaHCO3 is:

(1.52g / 2.372g) * 100 = 64.1%

7 0

4 years ago