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3 years ago

approximately how many grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution

1 answer:

4 0

3.00 M means 3.00 mol/L. So if you want 500 mL=0.5 L, it needs 3.00 mol/L * 0.5 L = 1.50 mol. For NaCl, the molar mass is 23+35.5=58.5 g/mol. So the amount of NaCl needed is 1.50*58.5=87.75 g. So the answer is 87.75 grams.

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The Sun heats land and water, how is this related to wind and ocean currents?

Yuki888 [10]

Answer:

"The sun warms up parts of the oceans. Warm waters rise just like warm air rises. So, as the warmer ocean waters begin to rise in a particular area, the cooler ocean waters from a different area will move in to replace the warmer ocean waters, and this creates our ocean currents."

Explanation:

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8 0

2 years ago

Compare and contrast chemical change and physical change

ki77a [65]

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3 years ago

What volume (mL) of concentrated solution of magnesium chloride (9.00 M) must be diluted to 350. mL to make a 2.75 M solution of

zalisa [80]

$❤Answer$

Volume of 106.9 mL from the concentrated solution should be taken and diluted to 350 mL.

⠀

Main I'd On Indian Brainly Is - HeartCrush

$⚡ Explantion$

We can use the formula. 

c1v1 =c2v2

Where c1 is the concentration and v1 is volume of the concentrated solution. 

c2 is the concentration

and v2 is the volume of the diluted solution to be prepared

9.00 M x V1 = 2.75 M x 350 mL

V1 = 106.9 mL

Volume of 106.9 mL from the concentrated solution should be taken and diluted to 350 mL.

3 0

2 years ago

PLEASE HELP!! I REALLY NEED HELP

Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na): $\frac{22.99}{58.443}$ = .393

2(Cl): $\frac{35.453}{58.443}$ = .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: $\frac{78.196}{138.204}$ = .566 Step 3: (.566)(100)= 56.6%

2: C: $\frac{12.011}{138.204}$ = .087 Step 3: (.087)(100)= 8.7%

2: O: $\frac{47.997}{138.204}$ = .347 Step 3: (.347)(100) = 34.7%

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: $\frac{167.535}{231.531}$ = .724 Step 3: (.724)(100)= 72.4%

2: O : $\frac{63.996}{231.531}$ = .276 Step 3: (.276)(100) = 27.6%

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: $\frac{36.033}{92.094}$ = .391 Step 3: (.391)(100) = 39.1%

2: H: $\frac{8.064}{92.094}$ = .088 step 3: (.088)(100) = 8.8%

2: O: $\frac{47.997}{92.094}$ = .521 step 3: (.521)(100) = 52.1%

3 0

3 years ago

Identify the following redox reactions by type. Check all that apply. (a) Fe + H2SO4 → FeSO4 + H2 combination decomposition disp

EastWind [94]

Answer :

(a) displacement reaction

(b) combination reaction

(c) disproportionation reaction

(d) displacement reaction

Explanation :

(a) The given balanced chemical reaction is,

$Fe+H_2SO_4\rightarrow FeSO_4+H_2$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Fe) replace the less reactive element (H).

(b) The given balanced chemical reaction is,

$S+3F_2\rightarrow SF_6$

This reaction is a combination reaction in which the two reactants molecule combine to form a large molecule or single product.

(c) The given balanced chemical reaction is,

$2CuCl_2\rightarrow Cu+CuCl_2$

This reaction is a disproportionation reaction in which the chemical species gets oxidized and reduced simultaneously. It is also considered as a redox reaction.

(d) The given balanced chemical reaction is,

$2Ag+PtCl_2\rightarrow 2AgCl+Pt$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Ag) replace the less reactive element (Pt).

7 0

3 years ago