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4 years ago

sam bought 2 1/4 pounds of swiss cheese 1 1/2 pounds of cheddar at the deli how many pounds of cheese did sam buy altogether

Chemistry

2 answers:

statuscvo [17]4 years ago

8 0

2 1^/4 + 11^/2 = 9^/4 + 3^/2 = 9^/4 + 6^/4 = 15^/4 = 3 3^/4

Lostsunrise [7]4 years ago

4 0

Answer:

$\frac{15}{4}$ pounds of cheese Sam had bought altogether.

Explanation:

Mass of Swiss cheese bought by Sam = $2\frac{1}{4}=\frac{9}{4} pounds$

Mass of cheddar cheese bought by Sam = $1\frac{1}{2}=\frac{3}{2} pounds$

Total mas of cheese bought by Sam:

$\frac{9}{4} pounds+\frac{3}{2} pounds$

$=\frac{9\times 2+3\times 4}{4\times 2} pounds=\frac{30}{8} pounds=\frac{15}{4} pounds$

$\frac{15}{4} pounds=3\frac{3}{4}$ pounds

$\frac{15}{4}$ pounds of cheese Sam had bought altogether.

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A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

7 0

3 years ago

Calculate the mass (g) of 1.5 moles of Sulfur.

just olya [345]

Answer:

48 g S

Explanation:

Step 1: Define

Molar Mass of Sulfur (S) - 32.07 g/mol

Step 2: Use Dimensional Analysis

$1.5 \hspace{3} mol \hspace{3} S(\frac{32.07 \hspace{3} g \hspace{3} S}{1 \hspace{3} mol \hspace{3} S} )$ = 48.105 g S

Step 3: Simplify

We have 2 sig figs.

48.105 g S ≈ 48 g S

7 0

3 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0

3 years ago

If 0.380 mol of barium nitrate is allowed to react with an excess of phosphoric acid, how many moles of barium phosphate could b

Vladimir [108]

If 0.380 mol of barium nitrate is allowed to react with an excess of phosphoric acid, 0.127 moles of barium phosphate could be formed.

<h3>How to calculate number of moles?</h3>

The number of moles of a compound can be calculated stoichiometrically as follows:

Based on this question, the following chemical equation is given:

Ba (NO3)2+ H3PO4 → Ba3(PO4)2+ HNO3

The balanced equation is as follows:

3Ba(NO3)2 + 2H3PO4 → Ba3(PO4)2 + 6HNO3

3 moles of barium nitrate produces 1 mole of barium phosphate

Therefore, 0.380 moles of barium nitrate will produce 0.380/3 = 0.127moles of barium phosphate.

Learn more about stoichiometry at: brainly.com/question/9743981

7 0

2 years ago

What is the basic unit of all matter? O A. neutron O B. atom KDC. electron OD proton NUNE nucleus Reset Next

Salsk061 [2.6K]

Answer:

Atom

Explanation:

atom: The basic unit of matter; the smallest unit of an element, having all the characteristics of that element; consists of negatively-charged electrons and a positively-charged center called a nucleus.

7 0

3 years ago

Read 2 more answers